Question 1 For the components in the sequential circuit shown below, t is the propagation delay, tatup is the setup time, and told is the hold time. The maximum clock frequency (rounded off to the nearest integer), at which the given circuit can operate reliably, is MHz. Flip Flop 1 = 2 ns fpa = 2 ns %3D fpa - 3 ns D- tsetup = 5 ns CLK frold =1 ns IN Flip Flop 2 fpd -8 ns fsetup = 4 ns fhold = 3 ns

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Question 1
For the components in the sequential circuit shown below, t is the propagation delay,
teatup is the setup time, and tolg is the hold time. The maximum clock frequency (rounded
off to the nearest integer), at which the given circuit can operate reliably, is
MHz.
Flip Flop 1
= 2 ns
pa = 2 ns
%3D
fpa
- 3 ns
betup = 5 ns
thold
D-
%3D
CLK
=1 ns
IN
Flip Flop 2
fpd
-8 ns
setup
= 4 ns
fhold
= 3 ns
Question 2
The components in the circuit given below are ideal. If R = 2 kQ and C = 1 µF, the
-3 dB cut-off frequency of the circuit in Hz is
R
ww
HH
C
R
V(jo) o
ov,(jo)
20
2R
ww
Transcribed Image Text:Question 1 For the components in the sequential circuit shown below, t is the propagation delay, teatup is the setup time, and tolg is the hold time. The maximum clock frequency (rounded off to the nearest integer), at which the given circuit can operate reliably, is MHz. Flip Flop 1 = 2 ns pa = 2 ns %3D fpa - 3 ns betup = 5 ns thold D- %3D CLK =1 ns IN Flip Flop 2 fpd -8 ns setup = 4 ns fhold = 3 ns Question 2 The components in the circuit given below are ideal. If R = 2 kQ and C = 1 µF, the -3 dB cut-off frequency of the circuit in Hz is R ww HH C R V(jo) o ov,(jo) 20 2R ww
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