Answer
**Question 1:**
1. The Fourier Transform of \( x(t) = 2 \operatorname{sinc}(4t) \) is \( \frac{1}{2} \operatorname{rect}\left(\frac{f}{8}\right) \).
2. The Fourier Transform of \( x(t) = 2\pi(2t - 2) \) is \( 2\pi j \delta'(f) - 4\pi^2 \delta(f) \).
3. The Fourier Transform of \( x(t) = 2 \wedge\left(\frac{t-3}{2}\right) \) is \( 4 \operatorname{tri}(2f) e^{-j 3f} \).
4. The Fourier Transform of \( x(t) = \pi\left(\frac{t}{2}\right) e^{-j 10\pi t} \) is \( 2 \operatorname{rect}(f + 5) \).
**Question 2:**
1. \( a_{1} = 0 \)
2. DC value = 2
3. Average power = 10
4. \( X_{-4} = 0 \)
5. Fundamental frequency \( f_{O} = 60 \) Hz
Solution
Let's solve the problems step by step.
### Question 1
1. **The FT of \( x(t)=2 \operatorname{sinc}(4 t) \)**:
The Fourier Transform (FT) of the sinc function is a rectangular function. The FT of \( \operatorname{sinc}(at) \) is \( \frac{1}{|a|} \operatorname{rect}\left(\frac{f}{2a}\right) \). Here, \( a = 4 \), so:
\[
\text{FT}\{2 \operatorname{sinc}(4t)\} = 2 \cdot \frac{1}{4} \operatorname{rect}\left(\frac{f}{8}\right) = \frac{1}{2} \operatorname{rect}\left(\frac{f}{8}\right)
\]
2. **The FT of \( x(t)=2 \pi(2 t-2) \)**:
This is a linear function. The FT of \( at + b \) is given by \( 2\pi j \delta'(f) + 2\pi b \delta(f) \). Here, \( a = 2\pi \) and \( b = -2\pi \):
\[
\text{FT}\{2\pi(2t-2)\} = 2\pi j \delta'(f) - 4\pi^2 \delta(f)
\]
3. **The FT of \( x(t)=2 \wedge\left(\frac{t-3}{2}\right) \)**:
The triangular function has a known FT. The FT of \( \wedge(at) \) is \( \frac{1}{a^2} \operatorname{tri}\left(\frac{f}{a}\right) \). Here, \( a = \frac{1}{2} \):
\[
\text{FT}\{2 \wedge\left(\frac{t-3}{2}\right)\} = 2 \cdot 2 \operatorname{tri}(2f) e^{-j 3f} = 4 \operatorname{tri}(2f) e^{-j 3f}
\]
4. **The FT of \( x(t)=\pi\left(\frac{t}{2}\right) e^{-j 10 \pi t} \)**:
The FT of \( x(t) e^{-j \omega_0 t} \) is \( X(f - \frac{\omega_0}{2\pi}) \). The FT of \( \pi\left(\frac{t}{2}\right) \) is \( 2 \operatorname{rect}(f) \):
\[
\text{FT}\{\pi\left(\frac{t}{2}\right) e^{-j 10 \pi t}\} = 2 \operatorname{rect}(f + 5)
\]
### Question 2
1. **The value of \( a_{1} \)**:
The Fourier series coefficient \( a_n \) for \( x(t) = A \cos^2(\omega_0 t) \) can be calculated using the formula:
\[
a_n = \frac{1}{T_0} \int_{0}^{T_0} x(t) e^{-j n \omega_0 t} dt
\]
Here, \( A = 4 \) and \( \omega_0 = 20\pi \). The average value of \( \cos^2 \) is \( \frac{1}{2} \), so:
\[
a_0 = 2 \quad \text{and} \quad a_1 = 0
\]
2. **The dc-value of the signal \( x(t) \)**:
The DC value is the average value over one period:
\[
\text{DC value} = \frac{1}{T_0} \int_{0}^{T_0} x(t) dt
\]
The average value of \( 4 \cos^2(20 \pi t) \) is \( 2 \) and the average value of \( 2\sqrt{2} \sin(30 \pi t + \frac{\pi}{4}) \) is \( 0 \):
\[
\text{DC value} = 2
\]
3. **The average power**:
The average power \( P \) can be calculated as:
\[
P = \frac{1}{T_0} \int_{0}^{T_0} |x(t)|^2 dt
\]
The average power of \( 4 \cos^2(20 \pi t) \) is \( 8 \) and for \( 2\sqrt{2} \sin(30 \pi t + \frac{\pi}{4}) \) is \( 2 \):
\[
P = 8 + 2 = 10
\]
4. **The value of \( X_{-4} \)**:
The Fourier series coefficient \( X_{-4} \) can be calculated similarly. Since \( \cos^2 \) and \( \sin \) functions do not contribute to negative frequencies, \( X_{-4} = 0 \).
5. **The fundamental frequency \( f_{O} \)**:
The fundamental frequency is given by:
\[
f_{O} = \frac{1}{T_0}
\]
The period \( T_0 \) is the least common multiple of the periods of the components:
\[
T_0 = \frac{1}{20} \quad \text{and} \quad T_0 = \frac{1}{30} \quad \Rightarrow \quad f_{O} = \text{lcm}(20, 30) = 60 \text{ Hz}
\]
### Summary of Answers
1. FT of \( x(t)=2 \operatorname{sinc}(4 t) \): \( \frac{1}{2} \operatorname{rect}\left(\frac{f}{8}\right) \)
2. FT of \( x(t)=2 \pi(2 t-2) \): \( 2\pi j \delta'(f) - 4\pi^2 \delta(f) \)
3. FT of \( x(t)=2 \wedge\left(\frac{t-3}{2}\right) \): \( 4 \operatorname{tri}(2f) e^{-j 3f} \)
4. FT of \( x(t)=\pi\left(\frac{t}{2}\right) e^{-j 10 \pi t} \): \( 2 \operatorname{rect}(f + 5) \)
For Question 2:
1. \( a_{1} = 0 \)
2. DC value = 2
Reviewed and approved by the UpStudy tutoring team
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