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**Unformatted text preview: **11)
(12) The same
comments as for
part c) apply here. Figure 7: Sketch of problem 2 part d. Problem 3
For better explanation, we draw a system diagram. Figure 8: System diagram of problem 3.
where
x(t) = repP (rect( 2t
))
P (13) Note ellipses. These
need to be included
to obtain full credit.
... ... Figure 9: Plot of x(t). 5 This not really drawn to scale the
intervals where the signal has value
0 should be of the same width as
those where the signal has value 1. P
1
P
1
comb P ( sinc( f ))
P
2
2
P
1
1
= comb P (sinc( f ))
2
2 X (f ) = F{x(t)} = (14) Figure 10: Plot of X (f ). a)
P = 1ms, 1
P = 1kHz, fc = 4kHz.
Z (f ) = HLP (f )X (f ) Figure 11: 6 (15) n
)
(16)
fs
ω
Zd (ω ) = fs repfs [Z ( fs )]
(17)
2π
1 ωfs
1
ωfs
ωfs
1
ωfs
ωfs
= fs repfs [ δ (
) + (δ (
− 1000) + δ (
+ 1000)) −
(δ (
− 3000) + δ (
+ 3000))]
2 2π
π
2π
2π
3π
2π
2π
z [n] = z (nTs ) = z ( Note δ ( ωfs ) =
2π 2π
fs δ (ω ).Therefore Zd (ω ) = rep2π [πδ (ω ) + 2(δ (ω − 2000π
2000π
2
6000π
6000π
) + δ (ω +
)) − (δ (ω −
) + δ (ω +
))]
fs
f...

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