Assignment 5 Solution
ECSE 306-B Assignment 5 Solution Set
Problem 3: (d) We need to prove
2
x(t ) dt
x 2 (nTs ) , where Ts
Ts n
By definition, x(nTs)
x[n] , so we need
2
x(t ) dt
1 . 2 f0
x 2 [n] .
Ts n
We first need to understand the relationship between the CTFT and DTFT. Let’s define the CTFT ( x(t ))
X ( jw) and DTFT ( x[n])
X (e j ) .
xp(t)
x(t)
p(t)
p (t )
(t nTs ) n
In continuous time domain, x p (t )
x(t ) (t nTs)
x(nTs) (t nTs) . The CTFT of x p (t ) is
n
n
X p ( jw)
jwnTs
x(nTs )e
by using the CTFT transform table.
n
In discrete time domain, x (e j )
x[n]e
j n
by definition, and we substitute in x(nTs)
x[n]
n
X (e j )
j n
x(nTs )e
.
n
By comparing X p ( jw) and X (e j ) , we can see that if we let w in continuous time and discrete time domain, i.e. X p ( j
Ts
)
Ts
, the spectrums are the same
X (e j ) .
Having the above relations, the following proof can be shown easily. From the class notes, you can find the DTFT of x p (t ) has another representation 1 Ts n
X p ( jw)
1 Ts n
have X (e j )
2 )) . By using the relationship derived above w Ts 2 k 1 X ( j( )) X ( j ( )) if . Ts Ts Ts
X ( j (w k
Then recall the DT Parseval’s theorem, x[n]
2
n
1 2
1 2
2 j
X (e ) d 2
1 X(j ) d . Ts Ts
Now, let w
Ts
, we have
, then substitute in the equation above,
Ts
, we
1 2
2
1 X( j ) d Ts Ts
1 2 Ts 2
/ Ts / Ts
2
X ( jw) Ts dw
1 1 Ts 2
/ Ts / Ts
2
X ( jw) dw * .
Finally, recall the CT Parseval’s theorem, 2
1 x(t ) dt X ( jw) dw , 2 since we are told Ts 1/ f 0 , we know for band limited signal, the value of the spectrum is zero w0 1 w0 outside the interval [ w0 , w0 ] , and w0 2 f 0 . Ts , and no f0 2 f 0 w0 Ts 2 aliasing will occur. 2
So,
2
x(t ) dt
1 2
2
X ( jw) dw
1 2
2
/ Ts
X ( jw) dw ** . / Ts
1 2 x(t ) dt . Ts n Multiply both sides by Ts, we get the equation that we need to proof.
Compare equation * and **, we have
x[n]
2
Problem 4 : (a) Refer to class notes (b) We know from the fourier transform table that :
Q(jw) can be represented as shown below:
Q(jw)
2π/Ts
c) Let
when
be represented as shown below: X(jw)
w
where
, then from the fourier transform table:
Since we know that
, then
is sketched as follows:
1/Ts
0
w
Notice from the sketch above that there is no aliasing. d) We are now asked to recover the original signal x(t) from the sampled signal xQ(t)= x(t)q(t). There are several ways to do this, here is one way. LPF 1
LPF 2
+
Where LPF 1 is a low pass filter such that
the
is used to up-modulate the
frequency. Finally LPF 2 is also a low pass filter such that
. These 3 steps are
shown below starting from the initial signal XQ(jw):
0
0
w
w
0
w
0
w
Problem 3: (d) We need to prove
2
x(t ) dt
x 2 (nTs ) , where Ts
Ts n
By definition, x(nTs)
x[n] , so we need
2
x(t ) dt
1 . 2 f0
x 2 [n] .
Ts n
We first need to understand the relationship between the CTFT and DTFT. Let’s define the CTFT ( x(t ))
X ( jw) and DTFT ( x[n])
X (e j ) .
xp(t)
x(t)
p(t)
p (t )
(t nTs ) n
In continuous time domain, x p (t )
x(t ) (t nTs)
x(nTs) (t nTs) . The CTFT of x p (t ) is
n
n
X p ( jw)
jwnTs
x(nTs )e
by using the CTFT transform table.
n
In discrete time domain, x (e j )
x[n]e
j n
by definition, and we substitute in x(nTs)
x[n]
n
X (e j )
j n
x(nTs )e
.
n
By comparing X p ( jw) and X (e j ) , we can see that if we let w in continuous time and discrete time domain, i.e. X p ( j
Ts
)
Ts
, the spectrums are the same
X (e j ) .
Having the above relations, the following proof can be shown easily. From the class notes, you can find the DTFT of x p (t ) has another representation 1 Ts n
X p ( jw)
1 Ts n
have X (e j )
2 )) . By using the relationship derived above w Ts 2 k 1 X ( j( )) X ( j ( )) if . Ts Ts Ts
X ( j (w k
Then recall the DT Parseval’s theorem, x[n]
2
n
1 2
1 2
2 j
X (e ) d 2
1 X(j ) d . Ts Ts
Now, let w
Ts
, we have
, then substitute in the equation above,
Ts
, we
1 2
2
1 X( j ) d Ts Ts
1 2 Ts 2
/ Ts / Ts
2
X ( jw) Ts dw
1 1 Ts 2
/ Ts / Ts
2
X ( jw) dw * .
Finally, recall the CT Parseval’s theorem, 2
1 x(t ) dt X ( jw) dw , 2 since we are told Ts 1/ f 0 , we know for band limited signal, the value of the spectrum is zero w0 1 w0 outside the interval [ w0 , w0 ] , and w0 2 f 0 . Ts , and no f0 2 f 0 w0 Ts 2 aliasing will occur. 2
So,
2
x(t ) dt
1 2
2
X ( jw) dw
1 2
2
/ Ts
X ( jw) dw ** . / Ts
1 2 x(t ) dt . Ts n Multiply both sides by Ts, we get the equation that we need to proof.
Compare equation * and **, we have
x[n]
2
Problem 4 : (a) Refer to class notes (b) We know from the fourier transform table that :
Q(jw) can be represented as shown below:
Q(jw)
2π/Ts
c) Let
when
be represented as shown below: X(jw)
w
where
, then from the fourier transform table:
Since we know that
, then
is sketched as follows:
1/Ts
0
w
Notice from the sketch above that there is no aliasing. d) We are now asked to recover the original signal x(t) from the sampled signal xQ(t)= x(t)q(t). There are several ways to do this, here is one way. LPF 1
LPF 2
+
Where LPF 1 is a low pass filter such that
the
is used to up-modulate the
frequency. Finally LPF 2 is also a low pass filter such that
. These 3 steps are
shown below starting from the initial signal XQ(jw):
0
0
w
w
0
w
0
w
