Show that a necessary and sufficient condition for \(p(t)\) to be a real function of time is that its Fourier transform \(\breve{P}(\omega)\) have the symmetry property \(\breve{P}(-\omega)=\breve{P} *(\omega)\), which shows that the real part of \(\breve{P}(\omega)\) is an even function of \(\omega\) but that its imaginary part is an odd function of \(\omega\) Then show that a necessary and sufficient condition for \(q(t)\), as given by (12 7.2) also to be real is for \(\breve{P}(\omega) \breve{T}(\omega)\), and hence \(\widetilde{T}(\omega)\), to obey the same symmetry property, so that \(\breve{T}(-\omega)={T}^{*}(\omega) .\) Note that this argument establishes the evenness and oddness of the real and imaginary parts of a transfer function without using the delta function.

Recall the definition of the Fourier transform: \(P(\omega) = \int_{-\infty}^{\infty} p(t)e^{-i\omega t} dt\). We will use this definition to find the required symmetry property.

Suppose that p(t) is a real function of time, meaning \(p(t)=p^*(t)\). Now we can find the Fourier transform of p(t), P(ω): \(P(-\omega) = \int_{-\infty}^{\infty} p(t)e^{i\omega t} dt\) Since p(t) is real, we can write this as: \(P(-\omega) = \int_{-\infty}^{\infty} p^*(t)e^{i\omega t} dt = P^*(\omega)\). This is the desired symmetry property: \(P(-\omega) = P^*(\omega)\).

If a function P(ω) has the symmetry property \(P(-\omega) = P^*(\omega)\), let's find out if p(t) is also real. We can use the inverse Fourier transformation of P(ω): \(p(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} P(\omega)e^{i\omega t} d\omega\) To find the complex conjugate of p(t), we first find the complex conjugate of the integrand: \(P^*(\omega)e^{-i\omega t} = P(-\omega)e^{-i(-\omega)t} = P(-\omega)e^{i\omega t}\) Thus, \(p^*(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} P^*(\omega)e^{-i\omega t} d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} P(-\omega)e^{i\omega t} d\omega\) Now we make a change of variables, let \(-\omega = v\), then \(d\omega = -dv\) \(p^*(t) = -\frac{1}{2\pi} \int_{\infty}^{-\infty} P(v)e^{-iv t} dv = \frac{1}{2\pi} \int_{-\infty}^{\infty} P(v)e^{-iv t} dv\) As both p(t) and p*(t) have the same expression, \(p(t) = p^*(t)\). Thus, p(t) is a real function.

Now we need to show that to make q(t) real, P(ω)T(ω) should also have the required symmetry property. Using a similar approach as before, we can say that if: \(P(ω)T(ω)\) has symmetry property, then: \(P(-\omega)T(-\omega) = P^*(\omega)T^*(\omega)\) And as we already know that P(ω) has symmetry property: \(P^*(\omega)T^*(\omega) = P^*(\omega)T(-\omega)\) This means \(\breve{T}(-\omega)={T}^{*}(\omega)\). This equality ensures that the evenness and oddness of the real and imaginary parts of the transfer function hold without using the delta function.