Expression to represent alternating pulse train?
Since this is a periodic signal, we have to first represent the signal $x(t)$ in terms of its Fourier Series and then compute its Fourier transform



$$\text{Representing } x(t) \text{ in terms of its Fourier series }$$

Let us assume $\Delta=T$



$$T_0=2T$$

$$x(t)=\sum_{n=-\infty}^\infty  x_ne^{j2\pi \frac{n}{T_0}t}=\sum_{n=-\infty}^\infty  x_ne^{j\pi nt}$$

$$x_n=\frac{1}{T_0}\int_{\alpha}^{\alpha +T_0} x(t)e^{-j2\pi \frac{n}{T_0}t}dt=\frac{1}{2T}\int_{0T^-}^{2T^-} x(t)e^{-j2\pi \frac{n}{2T}t}dt$$

$$\frac{1}{2T}\int_{0T^-}^{T^-} \delta (t)e^{-j\pi \frac{n}{T}t}dt+\frac{1}{2T}\int_{1T^-}^{2T^-} -\delta (t-T)e^{-j\pi \frac{n}{T}t}dt$$

$$=\frac{1}{2T}(1)-\frac{1}{2T}e^{-j\pi \frac{n}{T}T}=\frac{1}{2T}(1-e^{-j\pi n})=\frac{1}{2T}(1-(-1)^n)$$
$$x(t)=\sum_{n=-\infty}^\infty \frac{1}{2T}(1-(-1)^n)e^{j\pi\frac{n}{T}t} \tag 1$$



$$\text{Determing Fourier Transform of } x(t) \text{ from its Fourier series representation}$$

Let $n=2k+1$, then (1) becomes
$$=&gt; x(t)=\sum_{k=-\infty}^\infty \frac{1}{2T}(2)e^{j\frac{(2k+1)}{T}\pi t}$$

$$X(jw)=\int_{-\infty}^\infty x(t)e^{-jwt}dt$$
$$=\int_{-\infty}^\infty \sum_{k=-\infty}^\infty \frac{1}{2T}(2)e^{j\frac{(2k+1)}{T}\pi t}e^{-jwt}dt$$

$$\frac{1}{T}\sum_{k=-\infty}^\infty \int_{-\infty}^\infty [1.e^{j\frac{(2k+1)}{T}\pi t}]e^{-jwt}dt$$

$$\text{ Since } F[1]=2\pi\delta (w) \text{ and } F[e^{j\Omega_0t}]=2\pi \delta(\omega-\Omega_0) \text{ Therefore}$$

$$X(jw)=\frac{1}{T}\sum_{k=-\infty}^\infty 2\pi \delta(w-\frac{(2k+1)}{T}\pi)$$

$$=\frac{2\pi}{T}\sum_{k=-\infty}^\infty  \delta(w-\frac{(2k+1)}{T}\pi) \tag 2$$

$$X(f)=\frac{2\pi}{T}\sum_{k=-\infty}^\infty  \delta(2\pi f-\frac{2(2k+1)}{2T}\pi)$$

$$\text{Since } \delta(ax)=\frac{1}{|a|}\delta(x)$$

$$X(f)=\frac{2\pi}{|2\pi|T}\sum_{k=-\infty}^\infty  \delta( f-\frac{(2k+1)}{2T})$$

$$=\frac{1}{T}\sum_{k=-\infty}^\infty  \delta( f-\frac{(2k+1)}{2T})$$
Let $2k+1=m$
$$X(f)=\frac{1}{2T}\sum_{m=-\infty}^\infty (1-(-1)^m)\delta(f-\frac{m}{2T}) \tag 3$$