Solution to ECE 315 Test #8 F03
1.
Saying a signal has an infinite period and saying that it is aperiodic are really saying the same thing. (I also accepted
“non-periodic”, “not periodic”, “unperiodic” and “non-repeating” as correct.)
2.
The magnitude of the CTFT of any real-valued time-domain signal is always an even function of frequency and the
phase of the CTFT is always an odd function of frequency.
3.
Parseval’s theorem for the CTFT states that the signal energy of a signal is preserved in the transformation process.
4.
Below are two lists, one of time-domain functions and one of frequency-domain functions. By writing the number
of a time-domain function, match the frequency-domain functions to their inverse CTFT’s in the list of time-domain
functions. The match must be exact. If no exact match exists, just write “none”. Some of the functions may be written in
an unusual or unfamiliar form but still be correct, so check carefully.
Time Domain
Frequency Domain
4
−84 tri(12 f )
14
2
5 sinc(2( t + 2))
5
5
3δ ( 3t − 9)
 t
−7 sinc 2  
 12 
5 sinc(2( t − 2))
6
5 cos(200πt)
4
7
8
9
10
11
12
13
14
−96 sinc( 4 f )e j 2πf
1
 f
− comb − 
 8
2
13
1
 t + 5
2 tri

 10 
3δ ( t − 3)
−24[u( t + 1) − u( t − 3)]
−8 comb(−4 t)
 t − 4
9 rect 

 20 
 t + 10 
2 tri

 5 
−24 [u( t + 3) − u( t − 1)]
10 cos( 400πt )
]
11
−4 comb(8 t)
3
[
5 δ ( f − 200) + δ ( f + 200)
5
 f
rect   e − j 4 πf
 2
2
180 sinc(20 f )e − j 8πf
1
3
e − j 6πf
none
10 sinc 2 (5 f )e j10πf
The CTFT’s of these time-domain signals are
1
1
 f
 f
F
−4 comb(8 t) ←
→ − comb  = − comb −  , (an unshifted comb is an even function)
 8
 8
2
2
5
f
 
F
F
5 sinc(2( t + 2)) ←
→ rect   e j 4 πf
, 3δ ( 3t − 9) = δ ( t − 3) ←
→ e − j 6πf
 2
2
5
 t F
 f
F
−7 sinc 2   ←
→ −84 tri(12 f ) , 5 sinc(2( t − 2)) ←
→ rect   e − j 4 πf
 12 
 2
2
5
 t + 5 F
F
2
j10πf
5 cos(200πt) ←→ δ ( f − 100) + δ ( f + 100) , 2 tri
 ←→ 20 sinc (10 f )e
 10 
2
 t − 1 F
F
− j 2πf
3δ ( t − 3) ←
→ 3e − j 6πf , −24[u( t + 1) − u( t − 3)] = −24 rect 
 ←→ −96 sinc( 4 f )e
 4 
 t − 4 F
 f
 f
F
− j 8πf
−8 comb(−4 t) ←
→ −2 comb −  = −2 comb  , 9 rect 
 ←→180 sinc(20 f )e
 20 
 4
 4
 t + 1 F
 t + 10  F
2
j 20πf
j 2πf
2 tri
, −24[u( t + 3) − u( t − 1)] = −24 rect 
 ←→ −96 sinc( 4 f )e
 ←→10 sinc (5 f )e
 4 
 5 
F
10 cos( 400πt) ←
→ 5 δ ( f − 200) + δ ( f + 200)
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