Give step-by-step solution with explanation and final answer:Question 2: Recall that during the reconstruction of a band-limited signal z(t) from its samples z4[n], we used

an intermediate signal

Question

Give step-by-step solution with explanation and final answer:Question 2: Recall that during the reconstruction of a band-limited signal z(t) from its samples z4[n], we used

an intermediate signal
0
z(t) = Y_ a[n]d(t— nT)
n=—o0
and convolved it with the reconstruction filter h(t), where T' is the sampling period. We have also seen that if the
impulses &(t — nT’) are replaced with 7(t — nT) for any “narrow” signal r(t), we obtain an alternative intermediate
signal
oo
z(t) = Y alnlr(t— nT) =r(t) xz,(t) .
n=—o00
Therefore, the signal z(t), in effect, is filtered not by h(t), but by
he(t) = ho(t) x 7(t)

a) If (t) = e~ 7 u(t), and h(t) is the ideal reconstruction filter, i.c., h(t) = sinc (7+), find the continuous-time
Fourier transform H;(jQ2). Sketch the plot of |H,(j$)|.

b) Clearly, H,(j€) is not flat in the desired frequency interval —7 < Q < Z. For it to be flat, what should
H,(j) be? What is the corresponding h(t)?
Hint: For the last part, consider using the differentiation property of CTFT.

©) Another way to make sure HH, (jf) is flat is to use filtering in the discrete-time domain. In other words, one
can keep 7(t) and h,(t) as in part a, but insert a f

VIPSolutions · Accepted Answer

Let's break down solve each part step by step.

---

# **Part (a)**

**Given:**
- $ r(t) = e^{-\frac{10t}{T}}u(t) $
- $ h_r(t) = \text{sinc}\left(\frac{\pi t}{T}\right) $
- Find the continuous-time Fourier transform (CTFT) $\hat{H}_r(j\Omega)$ where:

$$
\hat{H}_r(j\Omega) = H_r(j\Omega) \cdot R(j\Omega)
$$

### **Step 1: Fourier Transform of $r(t)$**
The Fourier transform of $ r(t) = e^{-\alpha t} u(t) $ is:
$$
R(j\Omega) = \int_{}^{\infty} e^{-\alpha t} e^{-j\Omega t} dt = \frac{1}{\alpha + j\Omega}
$$
where $ \alpha = 10/T $.

So,
$$
R(j\Omega) = \frac{1}{\frac{10}{T} + j\Omega}
$$

---

### **Step 2: Fourier Transform of $h_r(t)$**
The ideal reconstruction filter is:
$$
h_r(t) = \text{sinc}\left(\frac{\pi t}{T}\right)
$$
whose Fourier transform is:
$$
H_r(j\Omega) = 
\begin{cases}
T, & |\Omega| \leq \frac{\pi}{T} \
, & \text{otherwise}
\end{cases}
$$

---

### **Step 3: Combined Frequency Response**
$$
\hat{H}_r(j\Omega) = H_r(j\Omega) \cdot R(j\Omega) = 
\begin{cases}
T \cdot \frac{1}{\frac{10}{T} + j\Omega}, & |\Omega| \leq \frac{\pi}{T} \
, & \text{otherwise}
\end{cases}
$$

---

### **Step 4: Sketch $|\hat{H}_r(j\Omega)|$**
$$
|\hat{H}_r(j\Omega)| = 
\begin{cases}
\frac{T}{\sqrt{\left(\frac{10}{T}\right)^2 + \Omega^2}}, & |\Omega| \leq \frac{\pi}{T} \
, & \text{otherwise}
\end{cases}
$$

- This function is **maximum** at $ \Omega =  $, then **decreases** as $ |\Omega| $ increases.
- It is not flat; it rolls off with frequency.

---

# **Part (b)**

### **Flat Response Requirement**
To have a flat frequency response $ |\hat{H}_r(j\Omega)| $ in $ -\frac{\pi}{T} \leq \Omega \leq \frac{\pi}{T} $:

$$
|\hat{H}_r(j\Omega)| = \text{constant}
$$

Given $ \hat{H}_r(j\Omega) = H_r(j\Omega) \cdot R(j\Omega) $, and $ H_r(j\Omega) $ is already ideal (rectangular):

- Therefore, $ R(j\Omega) $ must be **constant** (i.e., all-pass) in $ |\Omega| \leq \frac{\pi}{T} $.

### **What should $ r(t) $ be?**
The only signal whose Fourier transform is a constant (over all frequencies) is the **impulse**:
$$
r(t) = \delta(t)
$$

### **Corresponding Frequency Response**
$$
R(j\Omega) = 1
$$

### **Corresponding $ h_r(t) $?**
$$
h_r(t) = \text{sinc}\left(\frac{\pi t}{T}\right)
$$

---

# **Part (c)**

### **Goal:** 
Insert a filter $ h_d[n] $ in the discrete domain to **compensate** for the non-flatness of $ R(j\Omega) $.

### **Method:**
You want the overall system (including the discrete-time filter) to be flat in $ |\Omega| \leq \frac{\pi}{T} $.

Let $ H_d(e^{j\omega}) $ be the **DTFT** of $ h_d[n] $.

- CTFT: $ \Omega $
- DTFT: $ \omega = \Omega T $

#### The overall frequency response should be flat:
$$
H_d(e^{j\omega}) \cdot R(j\Omega) = 1, \quad (|\omega| \leq \pi)
$$
$$
H_d(e^{j\omega}) = \frac{1}{R(j\Omega)} = \frac{1}{R(j\omega/T)}
$$

Recall:
$$
R(j\Omega) = \frac{1}{\frac{10}{T} + j\Omega}
$$
So,
$$
H_d(e^{j\omega}) = \frac{10/T + j\omega/T}{1} = 10/T + j\omega/T
$$

This is the frequency response.

### **Find $ h_d[n] $:**
$ H_d(e^{j\omega}) = 10/T + j(\omega/T) $

- The first term, $ 10/T $, is a scaled impulse.
- The second term, $ j\omega/T $, corresponds (by DTFT property) to the differentiation in time, i.e., the sequence $ n $.

But to be exact: the inverse DTFT of $ j\omega $ is $ n $ times a delta function, but since this is a linear function of $ \omega $, it suggests a **differentiator**.

**So:**
$$
h_d[n] = \frac{10}{T}\delta[n] + \frac{1}{T}\cdot d[n]
$$
Where $ d[n] $ is the impulse response of the discrete-time differentiator. For practical purposes, this can be approximated by $ d[n] = \delta[n] - \delta[n-1] $, or ideally, a better approximation of the differentiator in discrete time.

---

## **Summary of Answers**

### **(a)**
$$
\hat{H}_r(j\Omega) = 
\begin{cases}
\frac{T}{\frac{10}{T} + j\Omega}, & |\Omega| \leq \frac{\pi}{T} \
, & \text{otherwise}
\end{cases}
$$

### **(b)**
To be flat, $ R(j\Omega) = 1 $, so $ r(t) = \delta(t) $.

### **(c)**
$$
H_d(e^{j\omega}) = 10/T + j\omega/T
$$
Thus,
$$
h_d[n] = \frac{10}{T}\delta[n] + \frac{1}{T}d[n]
$$
where $ d[n] $ is the discrete-time differentiator (impulse response).

---

## **Final Answers**

1. **(a)** $ |\hat{H}_r(j\Omega)| = \frac{T}{\sqrt{(10/T)^2+\Omega^2}} $, which is not flat—sketch as a lowpass-like curve.
2. **(b)** $ r(t) = \delta(t) $ (impulse), so $ R(j\Omega) = 1 $.
3. **(c)** Use a discrete-time filter $ h_d[n] $ with frequency response $ H_d(e^{j\omega}) = 10/T + j\omega/T $, i.e., combine a scaled impulse and a differentiator.

VIPSolutions · Answer

# **Part (a) Theory**

The Fourier transform of a signal describes its frequency components. For $ r(t) = e^{-\frac{10t}{T}}u(t) $, the CTFT is derived using integration. The ideal reconstruction filter $ h_r(t) = \text{sinc}\left(\frac{\pi t}{T}\right) $ has a rectangular frequency response. Their product gives insight into the combined system behavior in frequency.

---

# **Part (b) Theory**

To achieve a flat frequency response in the desired range, the Fourier transform $ R(j\Omega) $ should be constant (i.e., all-pass). The only signal with this property is the impulse function $ \delta(t) $, which has a constant FT of 1. Thus, the ideal $ h_r(t) $ remains unchanged.

---

# **Part (c) Theory**

Inserting a filter $ h_d[n] $ in the discrete domain can compensate for non-flatness in the frequency response. The desired flat frequency response can be achieved by ensuring $ H_d(e^{j\omega}) \cdot R(j\Omega) = 1 $. This requires combining a constant and a differentiator in the discrete domain representation of the filter.

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