第3章 8 11 12 25 27
第4章 15 27 29 32 35
第5章 12 23 33 35 49
1.4
Identify the following as open- or closed-loop control.
a.Controlling the water height in a toilet tank
b.Actuation of street lights at 6 P.M.
c.Stopping a clothes dryer when the clothes are dry
d.Actuation of an ice maker when the supply of cubes is low
識別以下作為開迴路或閉迴路控制。
a.控制馬桶水箱中的水高度
答;開迴路,通常是用水閥控制,系統不會知道現在的水位多少
b.在6:00 PM時路燈的點亮
答:開迴路,系統不會知道燈是否已經正確點亮,僅執行點亮的動作
c.當衣物乾燥時打開乾衣機
答:開迴路,乾衣機並不知道是否真的乾衣即打開
d.當製冰機的模具中的感測元件處於LOW的時候供水
答:閉迴路,因為有感測元件提供系統是否供水
1.5
A potentiometer has a transfer function of 0.1 V/deg.
Find the pot’s output if theinput is 45°.
電位器具有0.1V / deg的轉換函數。 如果輸入為45°,則找出其輸出。
答 : TF = output / input , 0.1 = 45 / input ,input = 450 v
1.6
A potentiometer has a transfer function of 0.05 V/deg.
Find the pot’s output if theinput is 89°.
電位器具有0.05V / deg的轉換函數。 如果輸入為89°,則找出其輸出。
答 : TF = output / input , 0.05 = 89 / input ,input = 445 v
1.7
A motor was measured to rotate (unloaded) at 500 rpm with a 6-V input and 1000 rpm with a 12-V input. What is the transfer function (steady state) for the unloaded motor?
A motor was measured to rotate (unloaded) at 500 rpm with a 6-V input and 1000 rpm with a 12-V input. What is the transfer function (steady state) for the unloaded motor?
測量電動機(卸載)在6V輸入下旋轉500rpm,以12V輸入旋轉1000rpm。 空載下此電動機的轉換函數(穩態)是什麼?
答:TF = output/input,6v * TF = 500rpm,12v * TF =1000rpm,TF = 83 rpm/v
1.8
in a certain system, an electric heating element was found to increase the temper-ature of a piece of metal 10° for each ampere of current. The metal expands 0.001in./deg and pushes on a load sensor which outputs 1 V/0.005 in. of compression.
a.Find the transfer functions of the three components and draw the block diagram.
b.Calculate the overall transfer function of this system.
在某個系統中,某個電加熱元件對每安培電流增加溫度10°。 此金屬片係數0.001in./deg,且傳輸每 1 V / 0.005 in.到負載感測器。
a。查找三個組件的傳遞函數並繪製框圖。
b。計算此系統的總轉換函數。
答:
TF = 0.1A/10°C * 0.001 in./deg * 200 v/in. = 0.02 v/A
第2章 7 8 10 12 14 18
2.7
A certain microprocessor has a simple instruction set shown below.
INSTRUCTION SET
Op-codeExplanation
76 Halt the microprocessor.
C6* Add next byte to accumulator.
D6* Subract next byte from accumulator.
3C Increment the accumulator.
3D Decrement the accumulator.
3E* Move the next byte into the accumulator.
*These instructions use two bytes.
What number would be in the accumulator after the program shown below was run?
Address Op-code
001 3E
002 05
003 D6
004 02
005 3C
006 76
某個微處理器具有如下所示的簡單指令集。
指令集
76 停止微處理器。
C6 *將下一個字節添加到累加器。
D6 *從累加器下拉字節。
3C 增加累加器。
3D 減少累加器。
3E *將下一個字節移入累加器。
*這些指令使用兩個字節。
在下面顯示的程序運行後,累加器中會有什麼數字?
答: 05 >> 03 >> 04
2.8
Temperature values from –20°F to 120°F are input data for a microprocessor com-puter.
Are 8 bits sufficient? If so, what is the resolution?
從-20°F到120°F的溫度值是微處理器計算機的輸入數據。 8位元是否足夠?
如果是,分辨率是多少?
答 : 8bit = 1~255,所以-20~120是足夠的,也就是0.549°
如果機械需要,設定為1°也是可以的。
2.10
Serial data are sent at 1200 bps using the format of Figure 2.11,
with one stop bit.How long would it take to send 1000 bytes of data?
2.12
The binary data from the computer in a certain application are expected
to go from00000000 to only 00111111. These data are the input of a DAC.
The analog outputshould go 0-5 V. Find the DAC reference voltage
necessary to make this happen.
在二進位系統中從00000000到00111111.
這些數據是DAC的輸入。 模擬輸出應為0-5 V。
找到使此發生所必需的DAC參考電壓。
答: 0-5 = 63 * Verf / 256
Verf = 0-20V
2.14
The binary output of an ADC should have the range 00000000-11111111
corre-sponding to an input of 0-6 V. Find the necessary reference voltage.
ADC的二進制輸出應該具有範圍00000000-11111111,
對應於0-6 V的輸入。找到必要的參考電壓。
答: 0-6 = 256 * Verf/256
Verf = 0-6V
2.18
A program contains 150 instructions, and the average execution time
per instruc-tion is 2 μs. Find the sample rate of this program.
程序包含150個指令,每條指令的平均執行時間為2μs。
查找此程序的採樣率。
答:1/2μs=0.5Msps
第3章 8 11 12 25 27
1.8
in a certain system, an electric heating element was found to increase the temper-ature of a piece of metal 10° for each ampere of current. The metal expands 0.001in./deg and pushes on a load sensor which outputs 1 V/0.005 in. of compression.
a.Find the transfer functions of the three components and draw the block diagram.
b.Calculate the overall transfer function of this system.
在某個系統中,某個電加熱元件對每安培電流增加溫度10°。 此金屬片係數0.001in./deg,且傳輸每 1 V / 0.005 in.到負載感測器。
a。查找三個組件的傳遞函數並繪製框圖。
b。計算此系統的總轉換函數。
答:
TF = 0.1A/10°C * 0.001 in./deg * 200 v/in. = 0.02 v/A
第2章 7 8 10 12 14 18
2.7
A certain microprocessor has a simple instruction set shown below.
INSTRUCTION SET
Op-codeExplanation
76 Halt the microprocessor.
C6* Add next byte to accumulator.
D6* Subract next byte from accumulator.
3C Increment the accumulator.
3D Decrement the accumulator.
3E* Move the next byte into the accumulator.
*These instructions use two bytes.
What number would be in the accumulator after the program shown below was run?
Address Op-code
001 3E
002 05
003 D6
004 02
005 3C
006 76
某個微處理器具有如下所示的簡單指令集。
指令集
76 停止微處理器。
C6 *將下一個字節添加到累加器。
D6 *從累加器下拉字節。
3C 增加累加器。
3D 減少累加器。
3E *將下一個字節移入累加器。
*這些指令使用兩個字節。
在下面顯示的程序運行後,累加器中會有什麼數字?
答: 05 >> 03 >> 04
2.8
Temperature values from –20°F to 120°F are input data for a microprocessor com-puter.
Are 8 bits sufficient? If so, what is the resolution?
從-20°F到120°F的溫度值是微處理器計算機的輸入數據。 8位元是否足夠?
如果是,分辨率是多少?
答 : 8bit = 1~255,所以-20~120是足夠的,也就是0.549°
如果機械需要,設定為1°也是可以的。
2.10
Serial data are sent at 1200 bps using the format of Figure 2.11,
with one stop bit.How long would it take to send 1000 bytes of data?
串行數據以1200 bps使用圖2.11的格式發送,帶有一個停止位元。
發送1000字節的數據需要多長時間?
答: 1200 / [1000/8*9] = 0.88s 2.12
The binary data from the computer in a certain application are expected
to go from00000000 to only 00111111. These data are the input of a DAC.
The analog outputshould go 0-5 V. Find the DAC reference voltage
necessary to make this happen.
在二進位系統中從00000000到00111111.
這些數據是DAC的輸入。 模擬輸出應為0-5 V。
找到使此發生所必需的DAC參考電壓。
答: 0-5 = 63 * Verf / 256
Verf = 0-20V
2.14
The binary output of an ADC should have the range 00000000-11111111
corre-sponding to an input of 0-6 V. Find the necessary reference voltage.
ADC的二進制輸出應該具有範圍00000000-11111111,
對應於0-6 V的輸入。找到必要的參考電壓。
答: 0-6 = 256 * Verf/256
Verf = 0-6V
2.18
A program contains 150 instructions, and the average execution time
per instruc-tion is 2 μs. Find the sample rate of this program.
程序包含150個指令,每條指令的平均執行時間為2μs。
查找此程序的採樣率。
答:1/2μs=0.5Msps
第3章 8 11 12 25 27