# How to prove a train of sinc pulses in digital communicaton system are orthogonal to each other?

Consider a train of sinc pulses: $$\phi_n(t)= \frac{\sin(\omega_M(t-nT_s))}{\omega_M(t-nT_s)}\quad; n=0,\pm1,\pm2,\dots$$ $$\quad$$where,$$\quad T_s=\frac{\pi}{\omega_M}$$
Now ,in order to show sinc pulses are orthogonal we need to prove: $$\int_{-\infty}^{\infty}\phi_n(t)\phi_k(t)dt=T_s \delta_{nk} \quad \dots(1)$$ where, $$\delta_{nk}$$ is kronecker's delta.

So, i began doing it as follows: $$T_s=\frac{\pi}{\omega_M}=\frac{\pi}{2\pi f_M}=\frac{1}{2f_M}=\frac{1}{f_N} \quad \dots(2)$$ where, $$f_N$$ is the Nyquist frequency $$\phi_0(t)=\frac{\sin(\omega_Mt)}{\omega_Mt}=\frac{\sin(2\pi f_Mt)}{2\pi f_Mt}=sinc(2f_Mt)=sinc(f_Nt) \quad \dots(3)$$ Now, $$\mathscr{F}\{ sinc(f_Nt) \}=\frac{1}{f_N} rect(\frac{f}{f_N})$$ ,where $$rect$$ is a rectangular function centred at origin and having width= $$f_N$$ $$\implies \mathscr{F}\{ sinc(f_N(t-nT_s)) \}=\frac{1}{f_N} \exp(-i2\pi f n T_s) rect(\frac{f}{f_N}) \quad \dots(4)$$ Now we can write: $$\int_{-\infty}^{\infty}\phi_n(t)\phi_k(t)dt=\int_{-\infty}^{\infty} \{ \Phi_n(f) \circledast \Phi_k(f) \} df$$ $$=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\Phi_n(\tau) \Phi_k(f-\tau) d\tau df$$ $$=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp(-i2\pi \tau n T_s) \frac{1}{f_N} rect(\frac{\tau}{f_N}) \exp(-i2\pi (f-\tau) k T_s) \frac{1}{f_N} rect(\frac{f-\tau}{f_N}) d\tau df$$ $$=\int_{-\infty}^{\infty} \exp(-i2\pi \tau n T_s) \frac{1}{f_N} rect(\frac{\tau}{f_N}) \exp(i2\pi \tau k T_s) \{ \int_{-\infty}^{\infty} \exp(-i2\pi f k T_s) \frac{1}{f_N} rect(\frac{f-\tau}{f_N}) df \} d\tau \quad \dots(5)$$ The inner integral of $$(5)$$ can be simplified as: $$\int_{\tau -\frac{f_N}{2}}^{\tau +\frac{f_N}{2}} \frac{1}{f_N} \exp(-i2\pi f k T_s) df$$ $$=\frac{\exp(-i2\pi \tau k T_s) \sin(\pi k)}{\pi k} \quad \dots(6)$$ So, $$(5)$$ can be rewritten as: $$\int_{-\infty}^{\infty} \exp(-i2\pi \tau n T_s) \frac{1}{f_N} rect(\frac{\tau}{f_N}) \exp(i2\pi \tau k T_s) \frac{\exp(-i2\pi \tau k T_s) \sin(\pi k)}{\pi k} d\tau$$ $$=\frac{\sin(\pi k)}{\pi k} \int_{-\infty}^{\infty} \exp(-i2\pi \tau n T_s) \frac{1}{f_N} rect(\frac{\tau}{f_N})d\tau$$ $$=\frac{\sin(\pi k)}{\pi k} \frac{\sin(\pi n)}{\pi n} \quad \dots(7)$$ Now, $$(7)$$ is even equal to $$0$$ when $$k=2$$ and $$n=2$$

So,where i missed ? any help or suggestions please...

The idea of solving the integral in the frequency domain is good, but you made a mistake rewriting the integral. Note that

$$\int_{-\infty}^{\infty}\phi_n(t)\phi_k(t)dt\tag{1}$$

equals the Fourier transform of $$\phi_n(t)\phi_k(t)$$ evaluated at $$f=0$$. As you know, that Fourier transform is given by the convolution of the two individual Fourier transforms of $$\phi_n(t)$$ and $$\phi_k(t)$$, respectively:

$$\mathcal{F}\big\{\phi_n(t)\phi_k(t)\big\}=\int_{-\infty}^{\infty}\Phi_n(\xi)\Phi_k(f-\xi)d\xi\tag{2}$$

Evaluating $$(2)$$ at $$f=0$$ gives

$$\int_{-\infty}^{\infty}\phi_n(t)\phi_k(t)dt=\int_{-\infty}^{\infty}\Phi_n(\xi)\Phi_k(-\xi)d\xi=\int_{-\infty}^{\infty}\Phi_n(\xi)\Phi_k^*(\xi)d\xi\tag{3}$$

where the last equality in $$(3)$$ is true because $$\phi_k(t)$$ is real-valued. Eq. $$(3)$$ is just Parseval's theorem.

I'm sure you can continue from here and show that the right-hand side of $$(3)$$ equals zero for $$n\neq k$$.

Note that the integral you tried to compute equals the inverse Fourier transform of $$(\Phi_n\star\Phi_k)(f)$$ evaluated at $$t=0$$, i.e., it equals $$\phi_n(0)\phi_k(0)$$ which also satisfies

$$\phi_n(0)\phi_k(0)=\delta[n-k]\tag{4}$$

• Thank you so much sir – Suresh Jul 27 '20 at 13:11