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You can use the convolution theorem to convert the convolution to the product of Fourier transforms of the given signals. $$h[n]*x[n]=\mathcal{F}^{-1}\{X(\omega)H(\omega)\}$$

$$H(\omega)=\mathcal{F}\{h[n]\}=\delta(\omega-\frac{\pi}{2})$$ $$X(\omega)=\mathcal{F}\{x[n]\}=e^{-j\frac{\omega 3 \pi}{2}}X_r(\omega)$$ where $$X_r(\omega)=\left\{\begin{matrix} \frac{\sin(2\omega)}{\sin(\omega/2)} &\omega\neq 0 \\ 4 & \omega=0 \end{matrix}\right.$$

However since $X(\omega)\delta(\omega-\frac{\pi}{2})=X(\frac{\pi}{2})\delta(\omega-\frac{\pi}{2})$ and $X(\frac{\pi}{2})=e^{-j\frac{3\pi}{2}}\frac{\sin(2\pi/2)}{\sin(\pi/4)}=0$, the inverse Fourier transform and therefore the covolution is zero.

You can also assume $x[n]$ is the impulse response of an LTI system and $h[n]$ is an input signal. Since the complex exponential is eigenfunction of the system, the output whould be an scaled version of the input i.e. $Ah[n]$ where $A=X(\frac{\pi}{2})$ and $X$ is the FT of $x[n]$.

You can use the convolution theorem to convert the convolution to the product of Fourier transforms of the given signals. $$h[n]*x[n]=\mathcal{F}^{-1}\{X(\omega)H(\omega)\}$$

$$H(\omega)=\mathcal{F}\{h[n]\}=\delta(\omega-\frac{\pi}{2})$$ $$X(\omega)=\mathcal{F}\{x[n]\}=e^{-j\frac{3 \omega}{2}}X_r(\omega)$$ where $$X_r(\omega)=\left\{\begin{matrix} \frac{\sin(2\omega)}{\sin(\omega/2)} &\omega\neq 0 \\ 4 & \omega=0 \end{matrix}\right.$$

However since $X(\omega)\delta(\omega-\frac{\pi}{2})=X(\frac{\pi}{2})\delta(\omega-\frac{\pi}{2})$ and $X(\frac{\pi}{2})=e^{-j\frac{3\pi}{4}}\frac{\sin(2\pi/2)}{\sin(\pi/4)}=0$, the inverse Fourier transform and therefore the covolution is zero.

You can also assume $x[n]$ is the impulse response of an LTI system and $h[n]$ is an input signal. Since the complex exponential is eigenfunction of the system, the output whould be an scaled version of the input i.e. $Ah[n]$ where $A=X(\frac{\pi}{2})$ and $X$ is the FT of $x[n]$.

You can use the convolution theorem to convert the convolution to the product of Fourier transforms of the given signals. $$h[n]*x[n]=\mathcal{F}^{-1}\{X(\omega)H(\omega)\}$$

$$H(\omega)=\mathcal{F}\{h[n]\}=\delta(\omega-\frac{\pi}{2})$$ $$X(\omega)=\mathcal{F}\{x[n]\}=e^{-j\frac{3\omega}{2}}X_r(\omega)$$ where $$X_r(\omega)=\left\{\begin{matrix} \frac{\sin(2\omega)}{\sin(\omega/2)} &\omega\neq 0 \\ 4 & \omega=0 \end{matrix}\right.$$

However since $X(\omega)\delta(\omega-\frac{\pi}{2})=X(\frac{\pi}{2})\delta(\omega-\frac{\pi}{2})$ and $X(\frac{\pi}{2})=e^{-j\frac{3\pi}{2}}\frac{\sin(2\pi/2)}{\sin(\pi/4)}=0$, the inverse Fourier transform and therefore the covolution is zero.

You can also assume $x[n]$ is the impulse response of an LTI system and $h[n]$ is an input signal. Since the complex exponential is eigenfunction of the system, the output whould be an scaled version of the input i.e. $Ah[n]$ where $A=X(\frac{\pi}{2})$ and $X$ is the FT of $x[n]$.

You can use the convolution theorem to convert the convolution to the product of Fourier transforms of the given signals. $$h[n]*x[n]=\mathcal{F}^{-1}\{X(\omega)H(\omega)\}$$

$$H(\omega)=\mathcal{F}\{h[n]\}=\delta(\omega-\frac{\pi}{2})$$ $$X(\omega)=\mathcal{F}\{x[n]\}=e^{-j\frac{\omega 3 \pi}{2}}X_r(\omega)$$ where $$X_r(\omega)=\left\{\begin{matrix} \frac{\sin(2\omega)}{\sin(\omega/2)} &\omega\neq 0 \\ 4 & \omega=0 \end{matrix}\right.$$

However since $X(\omega)\delta(\omega-\frac{\pi}{2})=X(\frac{\pi}{2})\delta(\omega-\frac{\pi}{2})$ and $X(\frac{\pi}{2})=e^{-j\frac{3\pi}{2}}\frac{\sin(2\pi/2)}{\sin(\pi/4)}=0$, the inverse Fourier transform and therefore the covolution is zero.

You can also assume $x[n]$ is the impulse response of an LTI system and $h[n]$ is an input signal. Since the complex exponential is eigenfunction of the system, the output whould be an scaled version of the input i.e. $Ah[n]$ where $A=X(\frac{\pi}{2})$ and $X$ is the FT of $x[n]$.

You can use the convolution theorem to convert the convolution to the product of Fourier transforms of the given signals.

$$\mathcal{F}\{h[n]\}=\delta(\omega-\frac{\pi}{2})$$ $$\mathcal{F}\{x[n]\}=\left\{\begin{matrix} \frac{\sin(2\omega)}{\sin(\omega/2)} &\omega\neq 0 \\ 4 & \omega=0 \end{matrix}\right.$$

Hence, $$h[n]*x[n]=\mathcal{F}^{-1}\{\frac{\sin(2\pi/2)}{\sin(\pi/4)}\delta(\omega-\frac{\pi}{2})\}=0$$

You can also assume $x[n]$ is the impulse response of an LTI system and $h[n]$ is an input signal. Since the complex exponential is eigenfunction of the system, the output whould be an scaled version of the input i.e. $Ah[n]$ where $A=X(\frac{\pi}{2})$ and $X$ is the FT of $x[n]$.

You can use the convolution theorem to convert the convolution to the product of Fourier transforms of the given signals. $$h[n]*x[n]=\mathcal{F}^{-1}\{X(\omega)H(\omega)\}$$

$$H(\omega)=\mathcal{F}\{h[n]\}=\delta(\omega-\frac{\pi}{2})$$ $$X(\omega)=\mathcal{F}\{x[n]\}=e^{-j\frac{3\omega}{2}}X_r(\omega)$$ where $$X_r(\omega)=\left\{\begin{matrix} \frac{\sin(2\omega)}{\sin(\omega/2)} &\omega\neq 0 \\ 4 & \omega=0 \end{matrix}\right.$$

However since $X(\omega)\delta(\omega-\frac{\pi}{2})=X(\frac{\pi}{2})\delta(\omega-\frac{\pi}{2})$ and $X(\frac{\pi}{2})=e^{-j\frac{3\pi}{2}}\frac{\sin(2\pi/2)}{\sin(\pi/4)}=0$, the inverse Fourier transform and therefore the covolution is zero.

You can also assume $x[n]$ is the impulse response of an LTI system and $h[n]$ is an input signal. Since the complex exponential is eigenfunction of the system, the output whould be an scaled version of the input i.e. $Ah[n]$ where $A=X(\frac{\pi}{2})$ and $X$ is the FT of $x[n]$.