cara mengubah bilangan pecahan decimal ke binary options

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Cara mengubah bilangan pecahan decimal ke binary options twitter bettingexpert football

Cara mengubah bilangan pecahan decimal ke binary options

Bilangan Oktal. Sistem bilangan Oktal menggunakan 8 macam symbol bilangan berbasis 8 digit angka, yaitu 0 ,1,2,3,4,5,6,7. Position value system bilangan octal adalah perpangkatan dari nilai 8. Operasi Aritmetika pada Bilangan Oktal. Langkah-langkah penjumlahan octal :.

Pengurangan Oktal dapat dilaukan secara sama dengan pengurangan bilangan desimal. Operasi Aritmetika Pada Bilangan Hexadesimal. Penjumlahan bilangan hexadesimal dapat dilakukan secara sama dengan penjumlahan bilangan octal, dengan langkah-langkah sebagai berikut :. Langkah-langkah penjumlahan hexadesimal :. Pengurangan bilangan hexadesimal dapat dilakukan secara sama dengan pengurangan bilangan desimal.

Konversi Bilangan. Yaitu dengan cara membagi bilangan desimal dengan dua kemudian diambil sisa pembagiannya. Konversi bilangan Desimal ke Oktal. Yaitu dengan cara membagi bilangan desimal dengan 8 kemudian diambil sisa pembagiannya. Contoh :. Konversi bilangan Desimal ke Hexadesimal. Yaitu dengan cara membagi bilangan desimal dengan 16 kemudian diambil sisa pembagiannya. Konversi ke desimal. Yaitu dengan cara mengalikan masing-masing bit dalam bilangan dengan position valuenya.

Konversi ke Oktal. Dapat dilakukan dengan mengkonversikan tiap-tiap tiga buah digit biner yang dimulai dari bagian belakang. Konversi ke Hexademial. Dapat dilakukan dengan mengkonversikan tiap-tiap empat buah digit biner yang dimulai dari bagian belakang. Konversi ke Desimal.

Konversi ke Biner. Dilakukan dengan mengkonversikan masing-masing digit octal ke tiga digit biner. Konversi ke Hexadesimal. Dilakukan dengan cara merubah dari bilangan octal menjadi bilangan biner kemudian dikonversikan ke hexadesimal. Konversi dari bilangan Hexadesimal. Sebutkan dan jelaskan empat macam system bilangan! Konversikan bilangan berikut :. Konversi dari :. Francois's whish is granted. We can also rewrite your four choises as.

Because Francois specified that he needed an apple AND a banana, you will only satisfy his command by bringing both. Sebutkan dan jelaskan empat macam system bilangan! Konversikan bilangan berikut : a. Excess-3 binary-coded decimal XS-3 , also called biased representation or Excess-N, is a numeral system used on some older computers that uses a pre-specified number N as a biasing value. It is a way to represent values with a balanced number of positive and negative numbers.

In XS-3, numbers are represented as decimal digits, and each digit is represented by four bits as the BCD value plus 3 the "excess" amount :. The smallest binary number represents the smallest value. To encode a number such as , then, one simply encodes each of the decimal digits as above, giving , , The primary advantage of XS-3 coding over BCD coding is that a decimal number can be nines' complemented for subtraction as easily as a binary number can be ones' complemented; just invert all bits.

In addition, when the sum of two XS-3 digits is greater than 9, the carry bit of a four bit adder will be set high. This works because, when adding two numbers that are greater or equal to zero, an "excess" value of six results in the sum. Since a four bit integer can only hold values 0 to 15, an excess of six means that any sum over nine will overflow.

Adding Excess-3 works on a different algorithm than BCD coding or regular binary numbers. When you add two XS-3 numbers together, the result is not an XS-3 number. For instance, when you add 1 and 0 in XS-3 the answer seems to be 4 instead of 1. In order to correct this problem, when you are finished adding each digit, you have to subtract 3 binary 11 if the digit is less than decimal 10 and add three if the number is greater than or equal to decimal 10 thus causing the number to wrap. Boolean Algebra The logic we apply in studying digital systems is based on a definition of truth made popular by Greek philosopher Aristotle BC BC.

To Aristotle, truth had two possible values: true or false: nothing is ever both true and false at the same time; nothing ever contradicts itself; and everything is what it is. In fact, Aristotles system of truth devised four laws: 1. The law of excluded middle: which says that either A is true or not A is true.

The law of inference: which says that you can arrive at one fact from other facts. About years after Aristotles work, an English mathematician named George Boole turned Aristotle's logic into an algebra. This brief history is part of the reason you are studying digital systems with logic algebra today. Let's start our exploration of Boolean algebra with three scenarios. And through these three scenarios, we will demonstrate ten basic facts concerning Boolean algebra.

For our first scenario, imagine that my friend Francois wants to eat a fruit and asks you to bring him an apple OR a banana. What this also means is that Boolean numbers and binary numbers arent the same. Binary numerals is just a translation of the decimal system; Boolean algebra is about logic.

Anyway, lets continue to our second scenario. Now lets apply the same Francois situation to another operator: AND. This time Francois wants to make smoothie and asks you to bring him an apple AND a banana. You still have four choices and each one of them will affect Francoiss wish. Because Francois specified that he needed an apple AND a banana, you will only satisfy his command by bringing both.

For the AND operator, we use the symbol in Boolean algebra. Hence, we can rewrite your four choices as Axiom 5. Axiom 6. Axiom 7. Axiom 8. If you think of the AND operator as the multiplication operator in normal algebra, then you are right on the money. They are not the same, but they work the same way.

Now lets proceed to our third and final scenario. Imagine this time that Francois is allergic to apple and tells you that no matter what you do, do NOT bring apple into the house. In this final case you only have two choices which of course will affect Francois. Hence, we rewrite your two choices as Axiom 9. Axiom The ten choices you just made regarding the different situations of Francois are known as the basic facts or axiom if you want to be fancy of Boolean algebra.

Most textbooks write these facts as complementary pairs to show socalled duality. For mere mortals like you and me, this complementary to show duality business simply means that in Boolean algebra if a statement is true then its opposite is also true. Your professor may not like that I am telling you that complementary means opposite, but trust me thats all you need to know.

Below are the five opposite pairs. I call them complements just to stay on the good side of your professor. But feel free to call them opposites. To show you how this complement stuff works, I have rearranged the ten axioms in pair of opposites. As you will find out soon, the duality of the complements i. But -. Dont try to memorize the list. You see how easy it is? Theorem 1. Theorem 2.

Theorem 3. Theorem 4. Theorem 5. Theorem 6. AC Theorem 9. I know that was a long list, but its a good idea to at least read through it once. Nobody memorizes this thing. As you simplify more and more Boolean expressions, they will come to you. Its as I said: we derive all of them from the axioms. Now, let me show you an example where we use the theorems to simplify an expression.

Here are a few more examples. You can do them first and then check your work against mine for reinforcement. Solution: We will simplify the right side of the equation to make it look like the left side. The output of the NOT Boolean operator is true when the input is false, and the output is false when the input is true.

Hence, all NOT does is take an operand and negate it. A philosopher will tell us the absence of good does not necessitate the presence of evil. However that may be, our concern here is not meaning. Our concern is the statement. As we show in Table 1, the operand Good can be in two possible states: either it is present or it is not present.

As such, we must evaluate the statement for two possibilities. We show the complete evaluation in Table 2. Therefore we rewrite the data from Table 2 in Table 3 accordingly. Physical Implementation In order to apply the principles of Boolean algebra to create real machines that can think and make decisions, we have had to find ways to physically implement the logic operators AND, OR, NOT, etc.

To that end, modern day engineering uses transistor networks called logic gates. Hence, a logic gate is actually a group of transistors so arranged as to behave as a Boolean operator. From a circuit complexity perspective, the most basic logic gate is the NOT gate aka the Inverter. The NOT gate is made of two transistors, as shown in Figure 1. In theory a NOT gate is really just one transistor.

But in practice microchip manufacturers use a pair of transistors to construct the NOT gate. Transistors The use of transistors to build logic gates is quite modern. Before transistors we used other devices, such as vacuum tubes aka thermionic valves. And very soon we may use DNA, or some other abundant material. There are many types of transistors.

The dominance is due to how well CMOS performs in all the important categories: fabrication cost, packing density, loading capacity i. There is of course more to transistors than can be presented here; especially since transistors are used for more than just digital systems. And so we refer you to any good microelectronics textbook. Each of the constructions presents specific conveniences to designers.

If you are very new to digital systems design, you may not understand the importance of the figures below. Still, we include them in this article for the people who may need them. Logic The AND gate is an electronic circuit that performs logical conjunction. The output of the AND Boolean operator is true only when all the inputs are true. Otherwise, the output is false. Through a short and simple analysis you can determine what a statement that uses the word AND is truly saying.

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Below are the five opposite pairs. I call them complements just to stay on the good side of your professor. But feel free to call them opposites. To show you how this complement stuff works, I have rearranged the ten axioms in pair of opposites. As you will find out soon, the duality of the complements i. But -. Dont try to memorize the list. You see how easy it is? Theorem 1.

Theorem 2. Theorem 3. Theorem 4. Theorem 5. Theorem 6. AC Theorem 9. I know that was a long list, but its a good idea to at least read through it once. Nobody memorizes this thing. As you simplify more and more Boolean expressions, they will come to you. Its as I said: we derive all of them from the axioms. Now, let me show you an example where we use the theorems to simplify an expression.

Here are a few more examples. You can do them first and then check your work against mine for reinforcement. Solution: We will simplify the right side of the equation to make it look like the left side. The output of the NOT Boolean operator is true when the input is false, and the output is false when the input is true.

Hence, all NOT does is take an operand and negate it. A philosopher will tell us the absence of good does not necessitate the presence of evil. However that may be, our concern here is not meaning. Our concern is the statement. As we show in Table 1, the operand Good can be in two possible states: either it is present or it is not present.

As such, we must evaluate the statement for two possibilities. We show the complete evaluation in Table 2. Therefore we rewrite the data from Table 2 in Table 3 accordingly. Physical Implementation In order to apply the principles of Boolean algebra to create real machines that can think and make decisions, we have had to find ways to physically implement the logic operators AND, OR, NOT, etc. To that end, modern day engineering uses transistor networks called logic gates.

Hence, a logic gate is actually a group of transistors so arranged as to behave as a Boolean operator. From a circuit complexity perspective, the most basic logic gate is the NOT gate aka the Inverter. The NOT gate is made of two transistors, as shown in Figure 1. In theory a NOT gate is really just one transistor.

But in practice microchip manufacturers use a pair of transistors to construct the NOT gate. Transistors The use of transistors to build logic gates is quite modern. Before transistors we used other devices, such as vacuum tubes aka thermionic valves. And very soon we may use DNA, or some other abundant material. There are many types of transistors. The dominance is due to how well CMOS performs in all the important categories: fabrication cost, packing density, loading capacity i.

There is of course more to transistors than can be presented here; especially since transistors are used for more than just digital systems. And so we refer you to any good microelectronics textbook. Each of the constructions presents specific conveniences to designers.

If you are very new to digital systems design, you may not understand the importance of the figures below. Still, we include them in this article for the people who may need them. Logic The AND gate is an electronic circuit that performs logical conjunction. The output of the AND Boolean operator is true only when all the inputs are true. Otherwise, the output is false.

Through a short and simple analysis you can determine what a statement that uses the word AND is truly saying. The statement tells us that the existence of Water depends on the existence of two objects: Hydrogen and Oxygen. Now each of these two objects can be in two possible definitive states: either an object exists or it does not exist.

Consequently we have four possible conditions, which we list in Table 1 below by row. You can list the possibilities however you want; you will still end up with four different cases. Try it if you want. Once we list the possible input conditions, we can proceed by adding a column for water as in Table 2.

So for example, the first row for water must be false since the statement did not say we can produce water out of nothing. We fill all the rows in Table 3 below. Notice that water is shown to exist only when both Hydrogen and Oxygen exist. Our circuits in figures 1, 2 and 3, for example, use complementary metal-oxide semiconductor CMOS technology.

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Carousel Previous Carousel Next. Jump to Page. Search inside document. Penjumlahan Langkah-langkah penjumlahan octal : tambahkan masing-masing kolom secara desimal rubah dari hasil desimal ke octal tuliskan hasil dari digit paling kanan dari hasil octal kalau hasil penjumlahan tiap-tiap kolom terdiri dari dua digit, maka digit paling kiri merupakan carry of untuk penjumlahan kolom selanjutnya.

Penjumlahan Penjumlahan bilangan hexadesimal dapat dilakukan secara sama dengan penjumlahan bilangan octal, dengan langkah-langkah sebagai berikut : Langkah-langkah penjumlahan hexadesimal : - tambahkan masing-masing kolom secara desimal rubah dari hasil desimal ke hexadesimal tuliskan hasil dari digit paling kanan dari hasil hexadesimal kalau hasil penjumlahan tiap-tiap kolom terdiri dari dua digit, maka digit paling kiri merupakan carry of untuk penjumlahan kolom selanjutnya.

Contoh : D 4 Konversi dari system bilangan Oktal 1. Konversi dari : a. BC1 2A X 5. In XS-3, numbers are represented as decimal digits, and each digit is represented by four bits as the BCD value plus 3 the "excess" amount : The smallest binary number represents the smallest value. Documents Similar To tugas kelas sebelas. Troy Campbell. Anissa 'ninis' Nuramalina. Pranesty Trisia Larasati. Siti Norazah Maulad Spawi. Febri Eka Setyawan. Farida Far Away Freez. Fajar Azzay. Qief Mandrez. Abrianto Nugraha.

Panji Rizki Maulana. Siti Nur Ain Sahat. Atau nilai perantara satu-terkait sebagai bagian-carry-out 1-bit dan nilai setengah-bit lebih tinggi 3 bit berdasarkan pada adder-carry-out Nilai half-level yang lebih tinggi dan nilai lower-half membentuk sebuah 6 - potongan jumlah bit yang sesuai dengan irisan addend 6 bit. Benda dan kelebihan lainnya dari penemuan ini akan menjadi jelas bagi mereka yang ahli. Dalam seni mengingat deskripsi mode terbaik yang diketahui saat ini dalam melakukan penemuan dan penerapan industri dari perwujudan yang disukai seperti yang dijelaskan di sini dan seperti yang digambarkan pada gambar gambar.

Tujuan dan keuntungan dari penemuan ini akan terlihat dari uraian terperinci berikut ini bersamaan dengan gambar gambar yang ditambahkan. Gambar 1A-B adalah diagram blok skematik dari perwujudan bit pertama contoh dari carry-select CS Penambah yang sesuai dengan penemuan ini, dimana Gambar 1A menunjukkan penambah CS secara rinci dan Gambar 1B menunjukkan penambah CS dengan referensi yang digunakan dalam diskusi. FIG 2 adalah diagram skematik yang menggambarkan konstruksi internal penukar kombinatorial 3-bit Blok yang digunakan pada penambah CS.

FIG 3 adalah diagram waktu penambah CS pada FIG 1A-B yang menunjukkan perambatan sinyal melalui tahapan setelah sinyal digital yang mewakili variabel input ar E disajikan pada jaring input dan garis bawaan. FIG 4 adalah diagram blok skematis yang menggambarkan perwujudan alternatif dari penambah CS inventif yang mungkin lebih disukai pada aplikasi dimana teknologi yang mendasari memiliki penundaan kawat yang signifikan.

Pada berbagai figur Gambar, seperti referensi digunakan untuk menunjukkan elemen atau langkah serupa atau yang serupa. Uraian Lengkap Penemuan. Perwujudan yang disukai dari penemuan ini adalah pemotong CS pembawa Seperti yang digambarkan dalam berbagai gambar di sini, dan khususnya pada pandangan Gambar Perwujudan yang disukai 1A-B dari penemuan ini digambarkan oleh karakter referensi umum Penemuan ini memberikan penambah CS penambah kecepatan tinggi yang dibawa cepat 10 Ini menggunakan blok penambah terkecil 3-bit yang merupakan penambah cepat yang pada dasarnya memiliki dua gerbang-penundaan Kinerja dan pemilihan paralel carry secara rekursif dalam tahap 6-bit.

Secara bersamaan, dua perwujudan teladan 18 bit dari penambah CS 10 disajikan sebagai contoh di sini Keduanya beroperasi pada dua bina Ry bit atau lebih kecil kata addend untuk memberikan carry-out 1 bit dan kata jumah bit biner Untuk diskusi, angka bit dipandang sebagai tiga irisan 6-bit yang masing-masing mencakup yang lebih tinggi - Setengah irisan 3 bit dan irisan 3 bit yang lebih rendah Setidaknya satu blok penambah 3 bit digunakan untuk masing-masing irisan 3-bit dari kata addend Salah satu blok adder digunakan untuk yang terendah, setengah bagian bawah 3-bit Slice bit 0 - 2 dan pengaturan blok penambah 3-bit duplikat digunakan untuk bit bit yang lebih signifikan 3 - 17 Pada perwujudan pertama, sebelas blok penambah 3-bit total digunakan dan pada lima belas total kedua bit Blok penambah digunakan.

Meskipun penambah CS inventif 10 juga dapat diimplementasikan dalam perwujudan ukuran kata lain, misalnya untuk menangani ukuran kata bit atau bit, perancang yang saat ini lebih disukai adalah perangkat bit yang menggunakan kombinasi 3-bit Blok penambah Hal ini terutama mengatasi kekurangan perangkat prior art, yang biasanya dirancang untuk lingkungan sinyal Menggunakan kelipatan 8-bit Ini juga telah terbukti sangat mudah beradaptasi untuk digunakan dalam array multiprosesor chip tunggal, sehingga memungkinkan perwujudan penarik CS inventif 10 untuk berfungsi dengan sangat baik pada perangkat yang dibuat oleh pemberi kerja sekarang.

Gambar 1A-B adalah skematik Diagram blok dari perwujudan bit pertama teladan dari penambah CS 10 yang sesuai dengan penemuan ini. Gambar 1A menunjukkan penambah CS 10 secara rinci dan Gambar 1B menunjukkan penambah CS 10 dengan referensi yang digunakan dalam diskusi berikut.

Adder 10 menerima tiga input dan menyediakan dua output Input termasuk kata addend 18 bit pertama yang diberikan pada jaring masukan pertama 12 kata addend 18 bit kedua yang diberikan pada jaring masukan kedua 14 dan sebuah carry-in opsional 1-bit yang disediakan Pada garis pembawa 16 Keluarannya mencakup sebuah kata jumlah 18 bit yang diberikan pada hasil jaring 18 dan bantalan 1 bit yang diberikan pada saluran keluar Sebentar juga untuk gambar 4, dua perwujudan dari Penambah CS 10 yang dibahas di sini masing-masing memiliki tiga m Bagian ajor 22 24 26 yang mencakup pengaturan dari empat jenis blok penambahan 6 bit 27 a - d Blok tambahan 27 a tidak membawa barang yang berarti dapat menerima salah satu atau nol pada carry-in line 16 Jika sebuah Penerima CS 10 tidak akan pernah perlu menerima carry-in, contoh blok tambahan 27 d dapat digunakan sebagai pengganti blok tambahan 27 a Sebaliknya, blok tambahan 27 b khusus untuk perwujudan pada Gambar 1A-B tidak terisi , Yang berarti bahwa hard-kabel untuk menggunakan nilai carry-in nol di blok penambah urutan terendah Sebaliknya, blok tambahan 27 c adalah satu-sarat, yang berarti kabel terprogram untuk menggunakan satu nilai carry-in.

Di blok penambah dengan tingkat terendah Dan blok tambahan 27 d khusus pada perwujudan pada Gambar 4 juga tidak terisi, walaupun menggunakan pengaturan komponen internal yang berbeda yang dibahas saat ini. Dengan mengacu pada gambar, dapat diapresiasi bahwa bagian 22 dan blok tambahan 27 A adalah satu dan sama, dan bagian 24 26 keduanya termasuk addit Blok ion 27 b dan blok tambahan 27 c atau blok tambahan 27 d dan blok tambahan 27 c Melihat bagian 22 24 26 dan blok tambahan 27 a-d dengan cara ini menekankan aspek rekursif penarik CS inventif 10 Dibahas lebih lanjut saat ini.

Lanjutkan sekarang hanya untuk gambar 1A-B di sini bagian 22 24 26 mencakup sebelas blok penambah kombinatorial blok penambah kolektif 28 blok penambah individual 28 a-k, lima multiplekser 2-ke-1 4-bit 30 a - e dua Multiplekser 7-bit 2-ke-1 32 a-b dan inverter masing-masing 34 untuk setiap multiplekser 30 a - e 32 a - b. Dua jaring masukan 18 garis 12 14 terpisah dibagi menjadi tiga subnet 36 38 40 yang membawa 6- Bit slice dari addend pertama dan kedua pada setiap bagian 22 24 26 Jadi, bit 0 - 5 dari kedua addend tersebut dikirimkan ke bagian 22 bit 6 - 11 dari kedua addend tersebut dikirim ke bagian 24 dan bit dari kedua addend tersebut dikirim.

Ke bagian 26 seperti yang ditunjukkan The 1-bit carry-in pada carry-in line 16 jika tersedia, juga dikirim ke Bagian Pada bagian 22 bagian 6-garis subnet 36 terpisah dibagi menjadi dua subnet 3-baris 44 46 seperti yang ditunjukkan Subnet 44 memberikan tiga bit dengan urutan lebih rendah yaitu potongan 3 bit paling bawah dari 6 bit terendah Slice, di sini bit 0 - 2 dari keduanya ditambahkan ke blok penambah 28 a dan subnet 46 memberikan tiga bit orde tinggi yaitu potongan 3 bit yang lebih tinggi dari potongan 6 bit yang sama, di sini bit 3 - 5 keduanya ditambahkan ke Blok penambah 28 b dan blok penambah 28 c Pengangkut 1-bit yang disediakan pada saluran masuk 16 dikirim ke blok penambah 28 a dan blok penambah 28 b dan blok penambah 28 c memiliki input terprogram dari 1 atau 0, seperti yang ditunjukkan.

Demikian pula, pada bagian 24 bagian 6 baris subnet 38 dibagi terpisah menjadi dua subnet 3-baris 48 50 seperti ditunjukkan Dan pada bagian 26 bagian 6 garis subnet 40 dibagi terpisah menjadi dua 3- Garis subnet 52 54 seperti yang ditunjukkan Pada bagian 24 26, blok penambah 28 d - k semuanya memiliki input terprogram baik 1 atau 0, seperti yang ditunjukkan. Melihat bagian 22 24 26 collec Sekarang masing-masing blok penambah 28 memberi makan subnet 4-baris masing-masing subnet subnet masing-masing subnet 56 a - k masing-masing multiplekser 4-bit 30 a-e memberi makan subnet 4-baris masing-masing 58 a-e dan masing-masing Dua multiplekser 7-bit 32 a-b memberi makan subnet 7-baris masing-masing 60 a - b Pencapaian efektif semua ini, yang didiskusikan dari perspektif fungsional saat ini, adalah bagian 22 yang dimasukkan ke subnet 62 bagian 24 dimasukkan ke dalam Subnet 64 bagian 26 feed ke subnet 66 dan subnet ini 62 64 66 digabungkan ke hasil bersih Pertimbangan blok penambah 28 a Dua bagian 3 baris di subnet 44 memberi makan dengan nilai bit 0 - 2 tiga paling tidak signifikan Bit LSB, dan carry-in line 16 memberi makan dengan nilai carry-in 1-bit Kemudian feed subnet 4-baris 56 a dengan nilai 4-bit terdiri dari jumlah bit dari masing-masing addend dan sebuah Bit adder-carry-out.

Selanjutnya pertimbangkan blok penambah 28 b dan blok penambah 28 c Dua bagian 3 baris pada subnet 46 feed keduanya Dengan nilai bit 3 - 5 pada addends Alih-alih bekerja dengan nilai carry aktual, blok penambah 28 b terprogram untuk menggunakan nilai nol dan blok penambah 28 c terprogram untuk menggunakan satu nilai. Dengan cara ini, blok penambah 28 b dan blok penambah 28 c menghitung kedua kemungkinan secara paralel, masing-masing memberi makan subnet 4-baris 56 b dan subnet 56 c dengan nilai menengah 4 bit yang terdiri dari jumlah bit 3 - 5 yang mungkin dari masing-masing addend.

Multiplexer 30 a menerima nilai antara dari blok penambah 28 b dan blok penambah 28 c pada subnet 56 b dan subnet 56 c dan berdasarkan bit adder-carry-out pada subnet 56 a melalui inverter 34 melewati intermediate yang sesuai. Jadi, bagian 22 mengeluarkan nilai 7-bit dimana tiga bit berurutan berasal dari subnet 56 a dan empat bit orde tinggi berasal dari subnet 58 a.

Secara khusus, bagian 22 mengeluarkan potongan jumlah 6 bit dari irisan 6 bit yang sesuai, dalam hal ini bit 0 - 5 dari kedua addend, dan nilai carry-out 1-bit Jumlah potongan bit 6 bit masuk ke Subnet 62 dan menjadi bit 0 - 5 pada hasil akhir pada hasil bersih 18 dan nilai carry-out bagian 1 bit digunakan oleh bagian 24 Melanjutkan protokol half-half yang lebih rendah yang digunakan saat membahas irisan masukan, ini 6 - Potongan jumlah bit dapat dilihat sebagai termasuk irisan 3 bit yang lebih tinggi dan irisan 3 bit bagian bawah.

Sisanya blok penambah 28 d - k digunakan dalam pengaturan pasangan seperti blok penambah 28 b - c Blok penambah 28 d - e menangani bit 6 - 8 blok penambah 28 f - g menangani bit 9 - 11 blok penambah 28 jam - i menangani bit 12 - 14 dan blok penambah 28 j - k handle Bit 15 - 17 seperti yang ditunjukkan pada FIG 1A-B. Now pertimbangkan bagian 24 Ada blok penambah 28 d - e hitung jumlah bit yang mungkin dari masing-masing addend, dan kemudian multiplekser 32 melewati sub-hasil yang sesuai ke subnet 64 Berdasarkan nilai carry-out 1 bit dari bagian 22 Pada saat bersamaan, blok penambah 28 f - g menghitung jumlah bit yang mungkin dari masing - masing addend dan memberikan nilai antara ini ke multiplekser 30 b - c yang kemudian masing - masing Lulus satu kemungkinan, berdasarkan masing-masing bit adder-carry-out pada subnet 56 d - e dari mana yang benar dilewatkan oleh multiplekser 32 yang masih bergantung pada nilai carry-out 1 bit dari bagian 22 Jadi, bagian 24 juga menghasilkan nilai 7-bit, yang merupakan potongan jumlah 6 bit bit penjumlahan dari kedua addend dan nilai carry-out 1 bit.

Potongan bit 6 bit masuk ke subnet 64 dan menjadi bit 6 - 11 pada hasil akhir hasil net 18 dan nilai carry-out 1-bit digunakan oleh bagian 26 Sekali lagi, potongan angka 6 bit ini c Sebuah juga dilihat sebagai termasuk irisan 3-bit yang lebih tinggi setengah dan irisan 3-bit bagian bawah.

Teknik yang sama digunakan pada bagian 26 sekarang untuk menghitung nilai 7-bit yaitu potongan jumlah 6 bit. Bit dari kedua addend dan nilai carry-out 1-bit Jumlah potongan bit 6 bit masuk ke subnet 66 dan menjadi bit dalam hasil akhir pada hasil bersih 18 dan bagian 1-bit-carry - nilai adalah output pada garis carry out 20 Dan lagi, potongan jumlah 6 bit ini dapat dilihat sebagai termasuk irisan 3 bit yang lebih tinggi dan irisan 3-bit yang lebih rendah.

Gambar 2 adalah diagram skematik Yang menggambarkan konstruksi internal blok penambah kombinatorial 3-bit 28 Elemen utama di blok penambah 28 adalah jala masukan tujuh baris 68 susunan inverter 70 jaring masukan empat belas baris 71 dan bidang 72 gerbang, dan sebuah OR Pesawat 74 gerbang Gerbang dasarnya pada dasarnya konvensional dan dapat memiliki hingga empat masukan, jadi gerbang NAND dengan lebih banyak masukan dibangun dari beberapa gerbang masukan 4 DAN yang menghubungkan ke NAN D gate.

The input net 68 mencakup dua subnet 3 baris dan satu garis bawaan tujuh baris di semua yang menerima potongan 3 bit tertentu dari dua kata addend, dan bit carry-in bit carry-in dapat hard wired to either 0 or 1, as described hereinabove and in the case of adder block 28 a it will be the 1-bit carry-in provided on carry-in line 16 The inverter array 70 has seven inverters that connect to input net 68 and provide inverted values on seven inverter output lines These inverter output lines are combined with input net 68 to form a fourteen-line complemented input net 71 which feeds seven un-inverted and seven inverted input bits and carry-in values to the AND plane On the input side of the AND plane 72 the inputs to the NAND gates are connected to particular lines of the input net 71 as needed according to known Boolean equations for bit sums and look-ahead carry values This provides 59 outputs, which are grouped by the sum bit being computed, to the OR plane FIG 3 is a timing diagram of a CS adder 10 showing signal propagation through the stages after digital signals representing input variables are presented at the input nets 12 14 and the carry-in line 16 The topmost trace in the diagram shows a signal level transition at time zero time , and the timing and the signal levels at various points in the CS adder 10 are shown by the other graph traces.

The signal transition at the outputs of the 3-bit combinatorial adder blocks 28 a - k at subnets 56 a - k are shown in the next graph trace, labeled 3-bits This is time at 2 5 time units, and it reflects the computation delay time of a 3-bit combinatorial adder block The next lower trace, labeled 6-bits, shows the signal tra nsition of the 4-line subnets 58 a - e below the 4-bit multiplexers 30 a - e This is time at three time units The difference between time and time thus represents the time delay introduced by a 4-bit multiplexer, for example multiplexer 30 a.

The further lower trace, labeled bits, shows the signal transition of the 7-line subnet 60 a below the first 7-bit multiplexer 32 a This is time at four time units The difference between time and time thus represents the time delay of multiplexer 32 a.

It should be noted that the carry-out to bit - 12 becomes available at time not earlier, and accordingly the high-order 6-bits of the bit sum require another 7-bit multiplexer delay The bottom trace, labeled bits, therefore shows the signal transition at subnet 60 b of multiplexer 32 b This is time at 5 time units.

It should be understood that the different components of the CS adder 10 comprising the adder blocks 28 multiplexers 30 32 and inverters 34 are themse lves composed of basic gates and circuit elements as known in the art, and can have characteristic delay times according to their types The times shown in FIG 3 are therefore approximate, and are chiefly shown for the purpose of clarifying the operation of the bit CS adder FIG 4 is a schematic block diagram depicting an alternate embodiment of the inventive CS adder 10 that may be preferred in applications where the underlying technology has significant wire delay To reduce wire delay owing to shared input connections between 4-bit multiplexers in sections 24 26 this approach employs more adder blocks 28 to permit closer connection from the adder blocks to the 4-bit multiplexers 30 b and 30 d All other aspects of the construction and operation of the CS adder 10 however, can remain substantially the same as described above.

Summarizing, the inventive CS adder 10 handles a carry-in and provides a carry-out and is suitable for various word lengths particularly including eighteen bit words Extremely high speed is achieved using the approach of multiplexing between two possible carry results computed in the MSB adder blocks simultaneously, and then selected by a carry computation from the LSB adder block Adder sections are made recursively of smaller adder blocks Unlike the conventional approach, however, where smallest blocks are brought down to the 1-bit level, the inventive CS adder 10 employs a 3-bit smallest adder block in a novel and particularly efficient manner that provides extremely high speed basically two gate delays for the computation of the 3-bit results and carries-out, simultaneously in parallel.

Of course, in alternate embodiments of the inventive CS adder 10 other types of 3-bit adders can be employed in place of the 3-bit combinatorial adder blocks 28 described above In particular, 3-bit ripple carry adders can be used, without otherwise altering the structure It will also be apparent to those skilled in the art that, with appropriate modificatio ns, other known multiplexer types may alternatively be used in other embodiments of the CS adder While various embodiments have been described above, it should be understood that they have been presented by way of example only, and that the breadth and scope of the invention should not be limited by any of the above described exemplary embodiments, but should instead be defined only in accordance with the following claims and their equivalents.

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Best Trading Sites. Carry-lookahead carry-select binary adder US A. That which is claimed is. The present invention relates to integrated circuits, and more particularly to integrated circuits for performing arithmetic operations.

Binary summation i e addition is one of the most important arithmetic operations performed by general-purpose and application specific processor systems e g digital signal processors This is because arithmetic summing operations are essential not only for addition, but also for subtraction, multiplication and division since these operations typically include repetitive summation steps Accordingly, the speed of microprocessors and other general-purpose arithmetic processors are hea vily dependent on the speed of the adder circuits contained therein.

Early microprocessor systems made use of classical adder designs, such as the ripple adder of FIG 1, which is a reproduction of FIG 2 2 from the textbook by J Cavanagh, entitled Digital Computer Arithmetic, McGraw Hill, Inc , the disclosure of which is hereby incorporated herein by reference Ripple adders are simple in design, require little electrical power and are easy to implement using conventional hardware, however, they are typically slow in their operation This is because ripple adders have relatively long propagation paths extending from the least significant bit to the most significant bit position of the adder Thus, a carry signal C is propagated in a time proportional to the size of the adder and hence, the size of the binary operands being summed As will be understood by those skilled in the art, the sum S of two binary operands B1 and B2 of length N can be obtained using the following well known relat ionships.

Many attempts have been made to increase the speed of arithmetic operations performed by general-purpose processors, based on a strategy of reducing the delay associated with carry propagation One such attempt, commonly referred to as carry-lookahead , is based on the principle that the carry-in signals for one or more higher-order adder stages can be generated directly from the inputs to the preceding lower-order stages without waiting for the carry-in signals to ripple through those stages Adders designed using this technique are commonly referred to as carry-lookahead adders CLA An exemplary CLA, including circuitry for generating group-propagate and group-generate signals, is shown in FIG 2 FIG 2 is a reproduction of FIG 2 5 from t he aforementioned Cavanagh textbook.

As shown in FIG 2, a conventional CLA looks at corresponding bit groups of two binary operands and generates a carry-out signal to the next higher order bit groups while the addition of the corresponding bit groups is performed to derive a sum Thus, the generation of the carry-out signal occurs in parallel i e simultaneously with the generation of the sum bits The lookahead circuitry reduces the need for rippling through every bit position and can reduce processing time to a value substantially below N There is, however, an area penalty caused by the additional lookahead circuitry As will be understood by those skilled in the art, group propagate, group generate and the carry-out signal for a four-bit group can be provided by circuitry which performs the following logic functions.

Another known adder design for increasing the speed of binary summation is shown in FIG 3, which is a reproduction of FIG 2 10 from the aforementioned Cavanagh textbook This adder includes pairs of group adder stages, as shown One of each pair performs summation operations assuming a carry bit from the preceding stage and the other performs summation operations assuming the absence of a carry bit from the preceding stage Group propagate and group generate signals, not shown, are also generated to derive the group carry bits GC 0 GC 1 GC 2 GC 3 as shown The adder of FIG 3 is commonly referred to by the acronym CSLA, because it combines features of conventional carry-select and carry-lookahead adders.

The carry-lookahead adder of FIG 4 is disclosed in U S Pat No 4,,, entitled Optimally Partitioned Regenerative Carry Lookahead Adder, to Vo et al FIG 4 is a reproduction of FIG 5 from the Vo et al patent, which is hereby incorporated herein by reference FIG 4 shows a bit full adder 60 arranged in a cascaded ripple fashion with bit-0 adder 50 being the least significant bit LSB adder and bit adder 65 b eing the most significant bit MSB adder Each bit adder 61 includes a circuit for generating propagate and generate signals not shown to its respective lookahead carry generation block 67 Each lookahead block 67 is arranged in a cascaded fashion so as to accept a carry-in from the previous block and generate a carry-out to the next subsequent block.

The bit adders 61 are arranged in irregular groupings to reduce the time associated with the propagation of the carry from the LSB adder to the MSB adder The grouping sequence is arranged by length from bit to bit-0 as , with the smallest bit groupings being at the least significant and most significant bit positions However, because of the cascaded arrangement, the propagation of the carry must still proceed serially through the blocks As will be understood by those skilled in the art, the worst case propagation path extends from the second bit position reference 53 to the last bit position reference 54 The path includes bit stage 1, look ahead blocks 2 through 7 and bit stages 29 and 30 Accordingly, the adder of FIG 4 has a worst case delay of T 2B 6L 1B, where B is the bit stage delay and L is the lookahead block delay The speed of the Vo et al bit adder is therefore limited by the serial propagation of the carry through the 6 intermediate blocks.

Other attempts to design fast adders include the carry-skip adder disclosed in an article by A Guyot, B Hochet and J Muller, entitled A Way to Build Efficient Carry-Skip Adders, IEEE Transactions on Computers, Vol C, No 10, October These adders comprise simple ripple adders with a plurality of speed-up carry chains skip chains The skip chains provide the feature whereby a carry into a block of full-adder cells can be bypassed to the next high order block if all the bits to be added in the block are different i e if p i 1 for all the cells in the block.

The adder also comprises pairs of 8-bit ripple adders for performing summation of 8-bit groups of the bit binary operands to be summed To achieve the carry-in signals at 8-bit intervals, the adder uses overlapping groups of carry-propagate and carry-generate signals, generated at the second and third tree levels, hence the term redundant These overlapping groups are generated at the intermediate outputs of the carry-chains As will be understood by those skilled in the art, the use of carry-chains having intermediate outputs causes additional delay to the generation of the carry-in signals by providing additional loading to the higher level chains in the tree Moreover, by using carry-chains of uniformly 4-bit length, the critical paths associated with the summation of each of the 8-bit groups of the bit operands are of relatively nonuniform length Thus, the sum bits for each of the consecutive 8-bit groups are not generated in the same amount of time.

Accordingly, notwithstanding the above-mentioned adder designs, there continues to be a need for fast binary adders, which are scalable and which have uniform carry-propagation delay times for performing carry-select and for generating groups of sum bits.

It is therefore an object of the present invention to provide an adder for performing summation of binary operands at a high rate of speed. It is another object of the present invention to provide an adder which can be scaled to perform summation of binary operan ds of varying length. It is a further object of the present invention to provide an adder wherein the critical path delays associated with the summation of respective portions of the binary operands to be added are of relatively uniform duration.

It is still a further object of the present invention to provide a binary adder which can be highly integrated on a semiconductor substrate. However, in the preferred embodiment, the adder also incorporates a recursive feature for increasing the speed of summation In this embodiment, each of the first and second binary ripple adders are replaced by an adder which includes the carry-lookahead and carry-select features of the invention, but on a smaller scale Thus, in the recursive embodiment, each of the first and second binary adders comprises a plurality of carry-lookahead cells of varying length at a first binary adder level and a carry-lookahead cell at a second binary adder level, which is electrically connected to the outputs of the first binary adder level cells Accordingly, the outputs of the second binary adder level cell depend on the carry-propagate and carry-generate outputs from the first binary adder level cells.

The first and second binary adders also comprise a plurality of pairs of smaller ripple adders connected e g indirectly via a multiplexer to the outputs of the second binary adder level cell Each of these pairs of ripple adders generates sum bits corresponding to the summation of respective sub-portions of the first and second binary operands One of each pair of the smaller ripple adders assumes the presence of a binary carry input binary 1 and the corresponding other assumes the absence of a binary carry input binary 0 , as described above with respect to the non-recursive embodiment Accordingly, to perform the carry-select function, the cell at the second binary adder level generates a plurality of outputs to select those ripple adders which made the correct assumption.

For sake of clarity, it is helpful to conceptualize the first and second binary operands as being formed of a plurality of corresponding consecutively ordered bit groups first, second nth The bit groups for each operand extend from a least significant bit group LSBG to a most significant bit group MSBG Similarly, each bit group contains a plurality of consecutively ordered bits, extending from a least significant bit LSB to a most significant bit MSB In addition, groups of carry-propagate bits p 0 p 1 p 2 p n and carry-generate bits g 0,g 1,g 2 g n can be generated from each of the corresponding pairs of operand bit groups using known relationships Moreover, each of the carry-propagate carry-generate bit groups can be partitioned into corresponding subgroups 0,g 0 p 1,g 1 p 2,g 2 p 3,g 3 p 4,g 4 p 5,g 4 p 6,g 6 p 7,g 7 p 8 g 8 p n, g n It is at the subgroup level that the first plurality of carry-lookahead cells are arranged in monotonically increasing order, by length.

In particular, the adder of the present invention comprises a first and a second binary adder for performing binary summation of a second bit group of the first binary operand and a corresponding second bit group of the second binary operand, simultaneously However, the first binary adder performs the summation assuming a binary carry into the least significant bits of the corresponding second bit groups and the second binary adder performs the summation assuming the absence of a binary carry into the least significant bits Accordingly, each of the first and second binary adders generates a different set of sum bits As will be understood by those skilled in the art, only one of the sets of sum bits will represent the correct summation, and it is the task of the rest of the adder to select the sum bits from the binary adder which assumed the proper carry input.

Based on the carry-generate and carry-propagate outputs of the first plurality of carry-lookahead cells, the second level carry-lookahead cell selects either the first or the second binary adder, depending on which generated the correct sum bits As will be understood by those skilled in the art, the invention may also comprise a plurality of second level cells and one or more third level cells, etc if the summation of operands having a relatively large number of bits is desired For example, the invention can perform the summation of bit operands using eleven 11 first level cells, three 3 second level cells and two 2 third level cells The cells preferably comprise Manchester carry-chains A bit adder according to the present inve ntion is ample for performing summation of the bit mantissas of IEEE Standard double-precision floating-point numbers.

FIG 1 schematically illustrates a prior art ripple adder which includes full-adder elements. FIG 2 schematically illustrates a prior art carry-lookahead adder. FIG 3 schematically illustrates a bit prior art carry-select adder. FIG 6 schematically illustrates a prior art 4-bit Manchester carry-chain, having intermediate outputs.

FIG 7 schematically illustrates a carry-lockhead carry-selected hybrid adder according to one embodiment of the present invention. FIG 8 schematically illustrates a carry-locked carry-selected hybrid adder according to a preferred embodiment of the present invention.

FIG 9 schematically illustrates a 4-bit Manchester carry-chain having a single pair of carry-generate carry-propagate outputs, according to the present invention. The present invention will now be described more fully hereinafter with reference to the accompanying drawings, in which preferred embodiments of the invention are shown This invention may, however, be embodied in different forms depending on the particular configuration or layout of the adder and should not be limited to the embodiments set forth herein Rather, these embodiments are provided so that this disclosure will be thorough and complete, and will fully convey the scope of the invention to those skilled in the art Like numbers refer to like elements throughout.

Referring now to FIG 7, an adder for performing binary summation of a first bit binary operand Augend B1 55 0 and a seco nd bit binary operand Addend B2 55 0 , according to the present invention, will be described The bit adder 10 comprises a pair of bit ripple adders 12A, 12B for performing binary summation of a second bit group of the first binary operand B1 23 9 and a corresponding second bit group of the second binary operand B2 23 9 The first bit ripple adder 12a performs the summation assuming a binary carry i e binary 1 into the least significant bits of the corresponding second bit groups B1 23 9 B2 23 9 and the second bit ripple adder 12b performs the summation assuming the absence of a binary carry i e binary 0 into the least significant bits.

The adder 10 further comprises a first, second and third plurality of carry-lookahead cells 14A-C at a first level and a plurality of carry-lookahead cells 16A-C and 18A-B at a second and third level, respectively The cells preferably comprise Manchester carry chains, as shown in FIGS 9 and 10, described hereinbelow The first plurality of ca rry-lookahead cells 14a at the first level are arranged by length in monotonically increasing order from a least significant carry-lookahead cell 20 to a most significant carry-lookahead cell 24 As is well known to those having skill in the art, a monotonically increasing sequence is a sequence of successive terms a i a i 1 a n where a i a i 1 a n.

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Integer, misalnya, dapat direpresentasikan dalam 8-bit, bit, bit atau bit Anda, sebagai pemrogram, memilih bit-bit yang sesuai untuk bilangan bulat Anda Pilihan Anda akan Menerapkan batasan pada bilangan bulat yang dapat diwakili. Selain bit-length, integer dapat ditunjukkan dalam berbagai skema representasi, misalnya unsigned vs signed integer.

Integer unsigned 8-bit memiliki range 0 sampai , sedangkan 8- Integer bertanda bit memiliki kisaran antara sampai - keduanya mewakili nomor yang berbeda. Penting untuk dicatat bahwa lokasi memori komputer hanya menyimpan pola biner. Hal ini sepenuhnya tergantung pada Anda, sebagai pemrogram, untuk menentukan bagaimana pola ini Harus ditafsirkan Sebagai contoh, pola biner 8-bit B dapat diartikan sebagai unsigned integer 65 atau karakter ASCII A atau beberapa informasi rahasia yang hanya diketahui oleh Anda Dengan kata lain, Anda harus terlebih dahulu memutuskan bagaimana merepresentasikan sepotong Data dalam pola biner sebelum pola biner Masuk akal Penafsiran pola biner disebut representasi data atau pengkodean Selanjutnya, penting agar skema representasi data disepakati oleh semua pihak, yaitu standar industri perlu dirumuskan dan diikuti secara langsung.

Setelah Anda memutuskan representasi data Skema, kendala tertentu, khususnya, ketepatan dan jangkauan akan diberlakukan Oleh karena itu, penting untuk memahami representasi data untuk menulis program kinerja yang benar dan berkinerja tinggi. Batu Lembing dan Pengkalian Hieroglif Mesir. Huruf hieroglif Mesir di sebelah kiri Digunakan oleh orang Mesir kuno sejak BC Sayangnya, sejak tahun M, tidak ada yang bisa lagi membaca hieroglif Mesir kuno, sampai ditemukannya kembali Rosette Stone pada tahun oleh pasukan Napoleon saat invasi Napoleon ke Mesir di dekat kota Rashid Rosetta di Delta Nil.

Batu Rosetta yang tersisa bertuliskan sebuah dekrit di BC atas nama Raja Ptolemy V Keputusan tersebut muncul dalam tiga scri Poin teks atas adalah hieroglif Mesir Kuno skrip Demotik paruh tengah, dan Yunani Kuno terendah Karena pada dasarnya menyajikan teks yang sama di ketiga skrip, dan bahasa Yunani Kuno masih dapat dipahami, ini memberi kunci pada penguraian hieroglif Mesir.

Moral dari cerita ini kecuali Anda mengetahui skema pengkodeannya, tidak mungkin Anda bisa memecahkan kode data. Referensi dan gambar Wikipedia. Integer Representation. Integers adalah bilangan bulat atau nomor titik tetap dengan titik radix yang dipelihara paling lambat. Mereka memiliki representasi yang berbeda dan diproses secara berbeda misalnya mengambang. Unsigned Integers can represent zero and positive integers. Signed Integers can represent zero, positive and negative integers Three representation schemes had been proposed for signed integers.

Sign-Magnitude representation. You, as the programmer, need to decide on the bit-length and representation scheme for your integers, depending on your application s requirements Suppose that you need a counter for counting a small quantity from 0 up to , you might choose the 8-bit unsigned integer scheme as there is no negative numbers involved.

Unsigned integers can represent zero and positive integers, but not negative integers The value of an unsigned integer is interpreted as the magnitude of its underlying binary pattern. Example 1 Suppose that n 8 and the binary pattern is B the value of this unsigned integer is 1 2 0 1 2 6 65D. Example 2 Suppose that n 16 and the binary pattern is B the value of this unsigned integer is 1 2 3 1 2 12 D.

Example 3 Suppose that n 16 and the binary pattern is B the value of this unsigned integer is 0. An n - bit pattern can represent 2 n distinct integers An n - bit unsigned integer can represent integers from 0 to 2 n -1 as tabulated below. Signed Integers. Signed integers can represent zero, positive integers, as well as negative integers Three representation schemes are available for signed integers.

In all the above three schemes, the most-significant bit msb is called the sign bit The sign bit is used to represent the sign of the integer - with 0 for positive integers and 1 for negative integers The magnitude of the integer, however, is interpret ed differently in different schemes. In sign-magnitude representation. The most-significant bit msb is the sign bit with value of 0 representing positive integer and 1 representing negative integer.

The remaining n -1 bits represents the magnitude absolute value of the integer The absolute value of the integer is interpreted as the magnitude of the n -1 - bit binary pattern. Example 1 Suppose that n 8 and the binary representation is 0 B Sign bit is 0 positive Absolute value is B 65D Hence, the integer is 65D. Example 2 Suppose that n 8 and the binary representation is 1 B Sign bit is 1 negative Absolute value is B 1D Hence, the integer is -1D.

Example 3 Suppose that n 8 and the binary representation is 0 B Sign bit is 0 positive Absolute value is B 0D Hence, the integer is 0D. Example 4 Suppose that n 8 and the binary representation is 1 B Sign bit is 1 negative Absolute value is B 0D Hence, the integer is -0D.

The drawbacks of sign-magnitude representation are. There are two representations B and B for the number zero, which could lead to inefficiency and confusion. Positive and negative integers need to be processed separately. In 1 s complement representation.

Again, the most significant bit msb is the sign bit with value of 0 representing positive integers and 1 representing negative integers. The remaining n -1 bits represents the magnitude of the integer, as follows.

Example 1 Suppose that n 8 and the binary representation 0 B Sign bit is 0 positive Absolute value is B 65D Hence, the integer i s 65D. Example 2 Suppose that n 8 and the binary representation 1 B Sign bit is 1 negative Absolute value is the complement of B i e B D Hence, the integer is D. Example 4 Suppose that n 8 and the binary representation 1 B Sign bit is 1 negative Absolute value is the complement of B i e B 0D Hence, the integer is -0D.

Again, the drawbacks are. There are two representations B and B for zero. The positive integers and negative integers need to be processed separately. In 2 s complement representation. Example 2 Suppose that n 8 and the binary representation 1 B Sign bit is 1 negative Absolute value is the complement of B plus 1 i e B 1B D Hence, the integer is D.

Example 4 Suppose that n 8 and the binary representation 1 B Sign bit is 1 negative Absolute value is the complement of B plus 1 i e B 1B 1D Hence, the integer is -1Dputers use 2 s Complement Representation for Si gned Integers. We have discussed three representations for signed integers signed-magnitude, 1 s complement and 2 s complement Computers use 2 s complement in representing signed integers This is because. There is only one representation for the number zero in 2 s complement, instead of two representations in sign-magnitude and 1 s complement.

Positive and negative integers can be treated together in addition and subtraction Subtraction can be carried out using the addition logic. Because of the fixed precision i e fixed number of bits , an n - bit 2 s complement signed integer has a certain range For example, for n 8 the range of 2 s complement signed integers is to During addition and subtraction , it is important to check whether the result exceeds this range, in other words, whether overflow or underflow has occurred. The following diagram explains how the 2 s complement works By re-arranging the number line, values from to are represented contiguously by ignoring the carry bit.

Range of n - bit 2 s Complement Signed Integers. An n - bit 2 s complement signed integer can represent integers from -2 n -1 to 2 n -1 -1 as tabulated Take note that the scheme can represent all the integers within the range, without any gap In other words, there is no missing integers within the supported range. Decoding 2 s Complement Numbers. Check the sign bit denoted as S. If S 0 the number is positive and its absolute value is the binary value of the remaining n -1 bits.

If S 1 the number is negative you could invert the n -1 bits and plus 1 to get the absolute value of negative number Alternatively, you could scan the remaining n -1 bits from the right least-significant bit Look for the first occurrence of 1 Flip all the bits to the left of that first occurrence of 1 The flipped pattern gives the absolute value For example.

Big Endian vs Little Endian. Modern computers store one byte of data in each memory address or location, i e byte addressable memory An bit integer is, therefore, stored in 4 memory addresses. The term Endian refers to the order of storing bytes in computer memory In Big Endian scheme, the most significant byte is stored first in the lowest memory address or big in first , while Little Endian stores the least significant bytes in the lowest memory address.

Exercise Integer Representation. What are the ranges of 8-bit, bit, bit and bit integer, in unsigned and signed representation. Give the value of 88 0 1 and in 8-bit unsigned representation. Give the value of 88 -1 0 1 and in 8-bit 2 s complement signed representation. Give the value of 88 -1 0 1 and in 8-bit sign-magnitude representation. Give the value of 88 -1 0 1 and in 8-bit 1 s complement representation.

The range of unsigned n - bit integers is 0, 2 n - 1 The range of n - bit 2 s complement signed integer is -2 n-1 , 2 n-1 Floating-Point Number Representation. A floating-point number or real number can represent a very large 1 23 10 88 or a very small 1 23 10 value It could also represent very large negative number -1 23 10 88 and very small negative number -1 23 10 88 , as well as zero, as illustrated.

A floating-point number is typically expressed in the scientific notation, with a fraction F , and an exponent E of a certain radix r , in the form of F r E Decimal numbers use radix of 10 F 10 E while binary numbers use radix of 2 F 2 E. Representation of floating point number is not unique For example, the number 55 66 can be represented as 5 10 1 0 10 2 0 10 3 and so on The fractional part can be normalized In the normalized form, there is only a single non-zero digit befo re the radix point For example, decimal number can be normalized as 1 10 2 binary number B can be normalized as 1 B 2 3.

It is important to note that floating-point numbers suffer from loss of precision when represented with a fixed number of bits e g bit or bit This is because there are infinite number of real numbers even within a small range of says 0 0 to 0 1 On the other hand, a n - bit binary pattern can represent a finite 2 n distinct numbers Hence, not all the real numbers can be represented The nearest approximation will be used instead, resulted in loss of accuracy.

It is also important to note that floating number arithmetic is very much less efficient than integer arithmetic It could be speed up with a so-called dedicated floating-point co-processor Hence, use integers if your application does not require floating-point numbers. In computers, floating-point numbers are represented in scientific notation of fraction F and exponent E with a radix of 2, in the form of F 2 E Both E and F can be positive as well as negative Modern computers adopt IEEE standard for representing floating-point numbers There are two representation schemes bit single-precision and bit double-precision.

In bit single-precision floating-point representation. The most significant bit is the sign bit S , with 0 for positive numbers and 1 for negative numbers. The following 8 bits represent exponent E. The remaining 23 bits represents fraction F.

Normalized Form. Let s illustrate with an example, suppose that the bit pattern is 1 with. F In the normalized form the actual fraction is normalized with an implicit leading 1 in the form of 1 F In this example, the actual fraction is 1 1 1 2 -2 1 2 -3 1 D. The sign bit represents the sign of the number, with S 0 for positive and S 1 for negative number In this example with S 1 this is a negative number, i e -1 D.

In normalized form, the actual exponent is E so-called excess or bias This is because we need to represent both positive and negative exponent With an 8-bit E, ranging from 0 to , the excess scheme could provide actual exponent of to In this example, E 2D. Hence, the number represented is -1 2 2 -5 5D. De-Normalized Form. Normalized form has a serious problem, with an implicit leading 1 for the fraction, it cannot represent the number zero Convince yourself on this.

De-normalized form was devised to represent zero and other numbers. For E 0 the numbers are in the de-normalized form An implicit leading 0 instead of 1 is used for the fraction and the actual exponent is always Hence, the number zero can be represented with E 0 and F 0 because 0 0 2 0. We can also represent very small positive and negative numbers in de-normalized form with E 0 For example, if S 1 E 0 and F The actual fraction is 0 1 2 -2 1 2 -3 0 D Since S 1 it is a negative number With E 0 the actual exponent is Hence the number is -0 2 -4 4 10 which is an extremely small negative number close to zero.

In summary, the value N is calculated as follows. For 1 E , N -1 S 1 F 2 E These numbers are in the so-called normalized form The sign-bit represents the sign of the number Fractional part 1 F are normalized with an implicit leading 1 The exponent is bias or in excess of so as to represent both positive and negative exponent The range of exponent is to For E 0, N -1 S 0 F 2 These numbers are in the so-called denormalized form The exponent of 2 evaluates to a very small number Denormalized form is needed to represent zero with F 0 and E 0 It can also represents very small positive and negative number close to zero.

For E it represents special values, such as INF positive and negative infinity and NaN no t a number This is beyond the scope of this article. Example 1 Suppose that IEEE bit floating-point representation pattern is 0 Example 2 Suppose that IEEE bit floating-point representation pattern is 1 Example 3 Suppose that IEEE bit floating-point representation pattern is 1 Exercises Floating-point Numberspute the largest and smallest positive numbers that can be represented in the bit normalized formpute the largest and smallest negative numbers can be represented in the bit normalized form.

Repeat 1 for the bit denormalized form. Repeat 2 for the bit denormalized form. Largest positive number S 0 E F Smallest positive number S 0 E 1 F 00 00 Same as above, but S 1. Notes For Java Users. You can use JDK methods bits or bits to create a single-precision bit float or double-precision bit double with the specific bit patterns, and print their values For examples. The representation scheme for bit double-precision is similar to the bit single-precision.

The following 11 bits represent exponent E. The remaining 52 bits represents fraction F. The value N is calculated as follows. More on Floating-Point Re presentation. There are three parts in the floating-point representation. The sign bit S is self-explanatory 0 for positive numbers and 1 for negative numbers. For the exponent E , a so-called bias or excess is applied so as to represent both positive and negative exponent The bias is set at half of the range For single precision with an 8-bit exponent, the bias is or excess For double precision with a bit exponent, the bias is or excess The fraction F also called the mantissa or significand is composed of an implicit leading bit before the radix point and the fractional bits after the radix point The leading bit for normalized numbers is 1 while the leading bit for denormalized numbers is 0.

Normalized Floating-Point Numbers. In normalized form, the radix point is placed after the first non-zero digit, e, g 9 D 10 D 1 B 2 11B For binary number, the leading bit is always 1, and need not be represented explicitly - this saves 1 bit of storage.

In IEEE s no rmalized form. For single-precision, 1 E with excess of Hence, the actual exponent is from to Negative exponents are used to represent small numbers 1 0 while positive exponents are used to represent large numbers 1 0 N -1 S 1 F 2 E Take note that n-bit pattern has a finite number of combinations 2 n , which could represent finite distinct numbers It is not possible to represent the infinite numbers in the real axis even a small range says 0 0 to 1 0 has infinite numbers That is, not all floating-point numbers can be accurately represented Instead, the closest approximation is used, which leads to loss of accuracy.

The minimum and maximum normalized floating-point numbers are. Special Values. Zero Zero cannot be represented in the normalized form, and must be represented in denormalized form with E 0 and F 0 There are two representations for zero 0 with S 0 and -0 with S 1. Not a Number NaN NaN denotes a value that cannot be represented as real number e g 0 0 NaN is represented with Exponent of all 1 s E for single-precision and E for double-precision and any non-zero fraction.

Character Encoding. In computer memory, character are encoded or represented using a chosen character encoding schemes aka character set , charset , character map , or code page. It is important to note that the representation scheme must be known before a binary pattern can be interpreted E g the 8-bit pattern B could represent anything under the sun known only to the person encoded it.

A 7-bit encoding scheme such as ASCII can represent characters and symbols An 8-bit character encoding scheme such as Latin-x can represent characters and symbols whereas a bit encoding scheme such as Unicode UCS-2 can represents 65, characters and symbols. ASCII is originally a 7-bit code It has been extended to 8-bit to better utilize the 8-bit computer memory organization The 8th-bit was originally used for parity check in the early computers. ANSI American National Standards Institute aka Windows or Windows Codepage for Latin alphabets used in the legacy DOS Windows systems It is a superset of ISO with code numbers 80H to 9FH assigned to displayable characters, such as smart single-quotes and double-quotes A common problem in web browsers is that all the quotes and apostrophes produced by smart quotes in some Microsoft software were replaced with question marks or some strange symbols It it because the document is labeled as ISO instead of Windows , where these code numbers are undefined Most modern browsers and e-mail clients treat charset ISO as Windows in order to accommodate such mis-labeling.

Before Unicode, no single character encoding scheme could represent characters in all languages For example, western european uses several encoding schemes in the ISOx family Even a single language like Chinese has a few encoding schemes GB GBK, BIG5 Many encoding schemes are in conflict of each other, i e the same code number is assigned to different characters.

Unicode aims to provide a standard character encoding scheme, which is universal, efficient, uniform and unambiguous Unicode standard is maintained by a non-profit organization called the Unicode Consortium Unicode is an ISO IEC standard Unicode has two encoding schemes. The transformation between Unicode and UTF-8 is as follows.

UTF is a variable-length Unicode character encoding scheme, which uses 2 to 4 bytes UTF is not commonly used The transformation table is as follows. Same as UCS-2 - no encoding. Same as UCS-4, which uses 4 bytes for each character - unencoded. Endianess or byte-order For a multi-byte character, you need to take care of the order of the bytes in storage In big endian the most significant byte is stored at the memory location with the lowest address big byte first In little endian the most significant byte is stored at the memory location with the highest address little byte first For example, with Unicode number of 60A8H is stored as 60 A8 in big endian and stored as A8 60 in little endian Big endian, which produces a more readable hex dump, is more commonly-used, and is often the default.

Unicode text files could take on these formats. Formats of Text Files. Windows CMD Codepage. Character encoding scheme charset in Windows is called codepage In CMD shell, you can issue command chcp to display the current codepage, or chcp codepage-number to change the codepage.

Codepage Windows , is not exactly the same as Latin-1 It assigns code number 80H to 9FH to letters and punctuation, such as smart single-quotes and double-quotes A common problem in browser that display quotes and apostrophe in question marks or boxes is because the page is supposed to be Windows, but mislabelled as ISO Chinese Character Sets.

Worse still, there are also various chinese character sets, which is not compatible with Unicode. BIG5 for traditional chinese characters BIG5 also uses 2 bytes for each chinese character The most significant bit of both bytes are also set to 1 BIG5 is not compatible with GBK, i e the same code number is assigned to different character.

For example, the world is made more interesting with these many standards. Collating Sequences for Ranking Characters. A string consists of a sequence of characters in upper or lower cases, e g apple BOY Cat In sorting or comparing strings, if we order the characters according to the underlying code numbers e g US-ASCII character-by-character, the order for the example would be BOY apple Cat because uppercase letters have a smaller code number than lowercase letters This does not agree with the so-called dictionary order where the same uppercase and lowercase letters have the same rank Another common problem in ordering strings is 10 ten at times is ordered in front of 1 to 9.

Hence, in sorting or comparison of strings, a so-called collating sequence or collation is often defined, which specifies the ranks for letters uppercase, lowercase , numbers, and special symbols There are many collating sequences available It is entirely up to you to choose a collating sequence to meet your application s specific req uirements Some case-insensitive dictionary-order collating sequences have the same rank for same uppercase and lowercase letters, i e A a B b Z z Some case-sensitive dictionary-order collating sequences put the uppercase letter before its lowercase counterpart, i e A B C a b c Typically, space is ranked before digits 0 to 9 followed by the alphabets.

Collating sequence is often language dependent, as different languages use different sets of characters e g , , a, with their own orders. For Java Programmers. JDK 1 4 introduced a new package to support encoding decoding of characters from UCS-2 used internally in Java program to any supported charset used by external devices.

Example The following program encodes some Unicode texts in various encoding scheme, and display the Hex codes of the encoded byte sequences. For Java Programmers - char and String. Java methods that accept a bit char value does not support supplementary characters Methods that accept a bit int value support all Unicode characters in the lower 21 bits , including supplementary characters. Langkah-langkah penjumlahan octal :. Pengurangan Oktal dapat dilaukan secara sama dengan pengurangan bilangan desimal.

Operasi Aritmetika Pada Bilangan Hexadesimal. Penjumlahan bilangan hexadesimal dapat dilakukan secara sama dengan penjumlahan bilangan octal, dengan langkah-langkah sebagai berikut :. Langkah-langkah penjumlahan hexadesimal :. Pengurangan bilangan hexadesimal dapat dilakukan secara sama dengan pengurangan bilangan desimal. Konversi Bilangan.

Yaitu dengan cara membagi bilangan desimal dengan dua kemudian diambil sisa pembagiannya. Konversi bilangan Desimal ke Oktal. Yaitu dengan cara membagi bilangan desimal dengan 8 kemudian diambil sisa pembagiannya. Contoh :. Konversi bilangan Desimal ke Hexadesimal. Yaitu dengan cara membagi bilangan desimal dengan 16 kemudian diambil sisa pembagiannya.

Konversi ke desimal. Yaitu dengan cara mengalikan masing-masing bit dalam bilangan dengan position valuenya. Konversi ke Oktal. Dapat dilakukan dengan mengkonversikan tiap-tiap tiga buah digit biner yang dimulai dari bagian belakang. Konversi ke Hexademial.

Dapat dilakukan dengan mengkonversikan tiap-tiap empat buah digit biner yang dimulai dari bagian belakang. Konversi ke Desimal. Konversi ke Biner. Dilakukan dengan mengkonversikan masing-masing digit octal ke tiga digit biner. Konversi ke Hexadesimal. Dilakukan dengan cara merubah dari bilangan octal menjadi bilangan biner kemudian dikonversikan ke hexadesimal.

Konversi dari bilangan Hexadesimal. Sebutkan dan jelaskan empat macam system bilangan! Konversikan bilangan berikut :. Konversi dari :. Francois's whish is granted. We can also rewrite your four choises as. Because Francois specified that he needed an apple AND a banana, you will only satisfy his command by bringing both. If you think of the AND operator as the multiplication operator in normal algebra, then you are right on the money.

They are not the same, but they work the same way. Hence, we rewrite your two choices as. The ten choices you just made regarding the different situations of Francois are known as the basic facts or axiom if you want to be fancy of Boolean algebra.

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Squad, sudah baca artikel tentang Pengertian dan Contoh Bilangan Bulat? Buat yang belum menyimaknya, kamu bisa pelajari dulu ya sebelum kita melanjutkan materi tentang 3 jenis bilangan pecahan serta bagaimana cara menyederhanakan dan mengubah dari tiap bilangan pecahan tersebut.

Lalu timbul pertanyaan, bagaimana mengubah pecahan biasa menjadi pecahan campuran maupun sebaliknya? Pembilang dibentuk dari bilangan bulat pada pecahan campuran yang dikalikan dengan penyebut, lalu ditambahkan pembilang pada pecahan campuran. Pecahan yang dinyatakan dalam bentuk dibaca: koma di mana a dan b bilangan bulat. Misalnya: 1,2; 0,4; 3, Mengubah pecahan ke desimal dengan pembagian bersusun Sumber: tes. Mengubah penyebut pecahan menjadi 10, , , dst sesuai banyaknya angka di belakang koma seperti pada gambar berikut:.

Bagaimana cara mengubah pecahan ke persen dan permil, juga sebaliknya? Contoh Soal dan Pembahasan. Cara mengurutkan desimal, pecahan biasa, dan persen adalah dengan sama-sama mengubahnya menjadi pecahan biasa, kemudian menyamakan penyebutnya.

Jadi, urutan pecahan terkecil ke terbesar:. Bagaimana, Squad? Sudah mengerti ' kan , apa saja jenis-jenis bilangan pecahan dan bagaimana cara mengubahnya dari setiap jenis tersebut. As'ari A. Cara ini dapat dimodifikasi untuk melakukan konversi dari basis desimal ke basis angka berapa saja. Angka pembagi adalah 2 karena basis sistem bilangan tujuan adalah basis 2 biner. Jika basis sistem bilangan tujuan adalah basis yang lainnya, gantilah angka basis 2 pada cara ini dengan angka basis yang sesuai.

Sebagai contoh, jika basis tujuan adalah basis 9, gantilah angka basis 2 dengan 9. Hasil akhir akan langsung dalam bentuk bilangan basis tujuan. Metode 2 dari Mulailah dengan membuat tabel. Naikkan pangkatnya dengan 1 untuk setiap pangkat. Lengkapi tabel sampai Anda mendapatkan sebuah bilangan yang paling dekat dengan bilangan dari sistem bilangan desimal yang Anda hitung.

Carilah bilangan dengan pangkat terbesar dari bilangan pokok 2. Dari tabel itu, pilihlah bilangan terbesar yang sama atau lebih kecil dari bilangan yang akan dikonversi. Lanjutkan ke pangkat berikutnya yang lebih kecil dalam tabel itu. Dengan memakai bilangan baru itu 28 , lanjutkan menelusuri tabel dari kiri ke kanan sambil memeriksa bilangan-bilangan tersebut, apakah sama atau lebih kecil dari bilangan baru. Lanjutkan sampai Anda menemui bilangan yang sama atau lebih kecil dari Teruskan sampai mencapai akhir tabel.

Tulislah jawaban bilangan biner itu. Seharusnya Anda mendapatkan hasil Pengulangan cara ini dapat membuat Anda mengingat bilangan-bilangan pangkat dari bilangan pokok dua, sehingga Anda bisa melewatkan langkah 1. Program Calculator yang sudah ada di dalam sistem operasi bisa melakukan konversi ini untuk Anda, tapi sebagai programer, lebih baik Anda memulai dengan pengertian yang baik tentang cara kerja konversi.

Melakukan konversi dengan arah sebaliknya, yaitu dari sistem bilangan biner ke desimal, biasanya lebih mudah untuk dipelajari lebih dahulu. Seringlah berlatih melakukan konversi bilangan desimal ke biner agar lebih ahli.

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