Fourier Transform

Given: h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos 40 \pi t
Fourier transform of signal \frac{\sin (10 \pi t)}{\pi t} is given by

Now, impulse response
h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right )\cdot \left ( \frac{e^{+j40\pi t}+e^{-j40\pi t}}{2} \right )
Using property, e^{-j\omega _0t}x(t)\rightleftharpoons X(\omega +\omega _0)
Therefore, Fourier transform of impulse response

Cut-off frequencies,
\omega_L=30 \pi rad/sec
\omega_H=50 \pi rad/sec
Thus, output of the system =7 \sin 42 \pi t+4 \cos 45 \pi t

Fourier transform, F(\omega)=A \tau S a^{2}\left(\frac{\omega \tau}{2}\right)
\begin{aligned} \text { As } A=1, \tau=1 \\ H \omega) &=S a^{2}\left(\frac{\omega}{2}\right) \\ \left.F(\omega)\right|_{\text {peak }} &=F(0)=S a^{2}(0)=1 \end{aligned}
\because Peak value of sampling function occurs at \omega=0
Peak value =1

f(t) \rightleftharpoons F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t
The following informations are given about f(t) \rightleftharpoons F(\omega).
\begin{array}{l} (i) f(t)=f(-t)\\ (ii) \left.F(\omega)\right|_{\omega=0}=1\\ (iii) \frac{d F(\omega)}{d \omega}=-\omega F(\omega)\\ \text{From }(iii), \frac{d F(\omega)}{d \omega}+\omega F(\omega)=0 \end{array}
By solving the above linear differential equations, (by mathematics)
\begin{aligned} \ln F(\omega) &=-\frac{\omega^{2}}{2} \\ \Rightarrow \qquad\qquad F(\omega) &=K \cdot e^{-\omega^{2} / 2} \\ \text { Put } \omega=0, \qquad\qquad F(0) &=K \\ \Rightarrow \qquad\qquad 1 &=K \text { (from info. } \\ \text { From (iv), } \qquad\qquad F(\omega) &=e^{-\omega^{2} / 2} \\ \text{As we know}, e^{-a t^{2}}, a \gt 0 &\rightleftharpoons \sqrt{\frac{\pi}{a}} e^{-\omega^{2} / 4 a}\\ \text{At }a=\frac{1}{2}, \qquad\qquad\quad e^{-t^{2} / 2} &\rightleftharpoons \sqrt{\frac{\pi}{1 / 2}} \cdot e^{-\omega^{2} / 2}\\ f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \rightleftharpoons e^{-\omega^{2} / 2}=F(\omega)\\ \text { Thus, }\qquad\qquad f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \\ \text { At } t=0, \qquad\qquad f(0)&=\frac{1}{\sqrt{2 \pi}} \lt 1 \end{aligned}

x(t)\rightleftharpoons X(j\omega )=\frac{1}{(10+j\omega )^2}
By taking inverse Fourier transform,
\begin{aligned} x(t)&=te^{-10t}u(t) \\ \text{Now, } x(t)_{t=1}&= 1 \times e^{-10} \times 1=e^{-10}\\ \text{Thus, }|\ln (x(t))| &= |\ln (e^{-10})|=|-10|=10 \end{aligned}

The Fourier transform of
\begin{aligned} &\frac{2 \sin (t\tau /2)}{t}\rightarrow 2 \pi \text{rect}\left ( \frac{\omega }{\tau } \right ) \\ &\frac{\sin (2 \pi t)}{\pi t}\rightarrow \text{rect} \left ( \frac{\omega }{4 \pi} \right )\\ &\text{So, }\int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} e^{-j\omega t}dt=\text{rect}\left ( \frac{\omega }{4 \pi} \right )\\ &\text{Putting }\omega =0 \text{ in the above equation} \\ & \int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} dt=1\\ &2\int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} dt=2 \end{aligned}

Maximum possible frequency of x(t)(2000 \pi t)=f_1+f_2=5+1=6kHz

By observing X_1(j \omega ) and X_2(j \omega ), we can say that they are not conjugate symmetric. Since, the fourier transform is not conjugate symmetric the signal will not be real. So, x_1(t), x_2(t) are not real. Now the fourier transform of x_1(t) \cdot x_2(t) will be \frac{1}{2\pi}X_1(j \omega )*X_2(j \omega ) and by looking at X_1(j \omega ) and X_2(j \omega ), we can say that X_1(j \omega ) *X_2(j \omega ) will be conjugate symmetric and thus x_1(t) \cdot x_2(t) will be real.
By observing X_1(j \omega ) and X_2(j \omega ), we can say,
X_2( \omega )=X_1(- \omega )\overset{IFT}{\rightarrow}x_2(t)=x_1(-t)\;\;...(i)
Now, X_1(j \omega ) is real. Therefore, x_1(t) will be conjugate symmetric.
\begin{aligned} \text{i.e., } x_1(t)&=x_1^*(-t) \\ \downarrow t&= -t\\ x_1(-t)&= x_1^*(t)\\ \text{Now, }x_1(t)\cdot x_2(t) &= x_1(t)\cdot x_1(-t) \;\;(from \;(i))\\ &= x_1(t)\cdot x_1^*(t)\\ &= |x_1(t)|^2=\text{real function} \end{aligned}

\begin{aligned} \text{Since, }x(t) &=\left\{\begin{matrix} e^{j10t} &\text{for }|t|\leq 1 \\ 0& \text{for }|t|\gt 1 \end{matrix}\right. \\ \text{as, } X(\omega )&=\int_{-\infty }^{\infty } x(t)e^{-j\omega t}dt\\ \text{so, } X(\omega )&=\int_{-1}^{1 } e^{j10t}e^{-j\omega t}dt\\ &=\int_{-1}^{1 } e^{jt(10-\omega )}dt\\ &= \frac{1}{j(10-\omega )}[e^{j(10-\omega )}- e^{-j(10-\omega )}]\\ X(\omega ) &=\frac{2 \sin (\omega -10)}{(\omega -10)} \end{aligned}

For real even function x(t), the Fourier transform X(\omega ) is always real even. y(t) is a derivative of x(t) which is a real odd function because derivative of even function is an odd function and hence, Fourier transform Y(\omega ) is imaginary odd.

\Rightarrow \;\; Symmetrically located pole and zero.
\Rightarrow \;\; All pass filter.
\Rightarrow \;\; Constant magnitude
(-\infty \leq \omega \leq \infty )