文档介绍:第二章部分****题答案
. (d) .=.
(e) .=.D
(i) .=.
(j) C.=.
. (a) ==C
(e) .
=.=FE.
. (b) EA==
(e)
=.
=.
. (e) .=.
(j) C. =.
. (b) =
(f) =DB
补充: .=. ,ε<-
. (a)
C
+
. (b)
B
+
. (b)
C
+
. (c)
C
F B
E
+
B
.
-bit signed-magnitude
+
-bit two’s
-bit one’s
+
+
-
-
-
. (b)
C
+
(c)
C
+
no overflow
overflow
第章部分****题答案
. (a) (b) (c) (d) probably
(e) undefined (f) probably (g)
(h) probably
. (a) (b) (c) (d) probably
(e) undefined (f) probably (g)
(h) probably
. A CMOS inverting gate has fewer
transistors.
. table -
VNH=VOHmin - VIHmin= .-.=.V (IOH=-μA)
VNH=VOHmin - VIHmin= .-.=.V (IOH=-mA)
VNL=VILmax - VOLmax= .-.=.V (IOL=μA)
VNL=VILmax - VOLmax= .-.=.V (IOL=mA)
. According to figure X.
VNH=VOHmin - VIHmin= .-.=.V
VNL=VILmax - VOLmax= .-.=.V
. (d)
IOH=i-i=VOH/R-(-VOH)/R
=./-(-.)/
=.mA-.mA
=.mA
VCC
i
i
IO
R
R
IOL=i-i=VOL/R-(-VOL)/R
=./-(-.)/
=.mA-.mA
=-.mA
. The voltage drop of resistor is:
VCC
R
LED
+
-
V
VOL
VR=.--VOL=-.=.V
Therefore
R=.V