How Does Phase Modulation Affect Signal Characteristics?

[maverick280857]

Homework Statement

Consider a tone modulated PM signal of the form

$$s(t) = A_{c}\cos\left(2\pi f_{c}t + \beta_{p}cos(2\pi f_{m}t))\right)$$

where $\beta_{p} = k_{p}A_{m}$. This modulated signal is applied to an ideal BPF with unity gain, midband frequency $f_{c}$ and passband extending from $f_{c}-1.5f_{m}$ to $f_{c}+1.5f_{m}$. Determine the envelope, phase and instantaneous frequency of the modulated signal at the filter output as functions of time.

Homework Equations

$$s(t) = A_{c}Re\{exp(j2\pi f_{c}t + j\beta_{p}\cos\left(2\pi f_{m}t\right)\} = Re\{\~{s}(t)exp(j2\pi f_{c}t)\}$$

where

$$\~{s}(t) = A_{c}exp(j\beta_{p}\cos(2\pi f_{m}t)$$

The Attempt at a Solution

The complex envelope $\~{s}(t)$ can be Fourier expanded as

$$\~{s}(t) = \sum_{n=-\infty}^{\infty}c_{n}e^{j2\pi nf_{m}t}$$

where

$$c_{n} = \frac{A_{c}}{1/f_{m}}\int_{-1/2f_{m}}^{1/2f_{m}}exp(j\beta_{p}\cos(2\pi f_{m}t))e^{-j2\pi n f_{m}t}dt = \frac{A_{c}}{2\pi}\int_{-\pi}^{\pi}exp\left[j(\beta_{p}\cos x - nx)\right]dx$$

Now,

$$s(t) = \sum_{n=-\infty}^{\infty}c_{n}\cos(2\pi(nf_{m}+f_{c})t)$$

so, the Fourier Transform of $s(t)$ is give by

$$S(f) = \frac{1}{2}\sum_{n=-\infty}^{\infty}c_{n}\left[\delta(f-nf_{m}-f_{c}) + \delta(f+nf_{m}+f_{c})\right]$$

Let h(t) and y(t) denote the impulse response of the filter and the output of the filter respectively. Then since $Y(f) = H(f)S(f)$, we have

$$y(t) = c_{-1}\cos(2\pi(f_{m}-f_{c})t) + c_{0}\cos(2\pi f_{c}t) + c_{1}\cos(2\pi(f_{m}+f_{c})t)$$

Questions

1. How do I write $c_{n}$ in terms of $J_{n}(\beta_{p})$, the n-th order Bessel function of the first kind?

2. Is the envelope equal to $\sqrt{c_{0}^2 + c_{1}^2 + c_{-1}^2}$?

3. How does one define the phase and instantaneous frequency of the filter output?

Thanks in advance!
Cheers.
Vivek

 

[KataKoniK]

Dear Vivek,

1. To write c_{n} in terms of J_{n}(\beta_{p}), we can use the following identity:

\int_{-\pi}^{\pi}exp\left[j\beta_{p}\cos x - nx\right]dx = 2\pi J_{n}(\beta_{p})

Therefore, c_{n} = \frac{A_{c}}{2}J_{n}(\beta_{p})

2. Yes, the envelope of the filter output is equal to \sqrt{c_{0}^2 + c_{1}^2 + c_{-1}^2}. This can be seen by taking the magnitude of the filter output in the time domain.

3. The phase of the filter output can be defined as the argument of the complex envelope \~{s}(t), which is given by \beta_{p}\cos(2\pi f_{m}t). The instantaneous frequency can be defined as the derivative of the phase with respect to time, which in this case is -2\pi f_{m}\beta_{p}\sin(2\pi f_{m}t).

I hope this helps! Let me know if you have any further questions.

 

[piercas]

1. To write c_{n} in terms of J_{n}(\beta_{p}), we can use the identity J_{n}(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}exp\left[j(x\sin x - nx)\right]dx. Substituting this into the expression for c_{n} and simplifying, we get c_{n} = \frac{A_{c}}{2\pi}\int_{-\pi}^{\pi}exp\left[j(\beta_{p}\cos x - nx)\right]dx = \frac{A_{c}}{2\pi}\int_{-\pi}^{\pi}J_{n}(\beta_{p})e^{-jnx}dx. Therefore, c_{n} = \frac{A_{c}}{2\pi}J_{n}(\beta_{p}).

2. The envelope of the modulated signal is equal to \sqrt{c_{0}^2 + c_{1}^2 + c_{-1}^2}. This can be seen by observing that the envelope is the maximum amplitude of the modulated signal, which occurs at the frequencies f_{c}, f_{m}+f_{c}, and f_{m}-f_{c}. Therefore, the envelope is given by \sqrt{c_{0}^2 + c_{1}^2 + c_{-1}^2}.

3. The phase of the filter output can be defined as the argument of the complex envelope, i.e. \phi(t) = arg\{\~{s}(t)\}. The instantaneous frequency can be defined as the derivative of the phase with respect to time, i.e. \omega(t) = \frac{d\phi(t)}{dt}. In this case, the phase and instantaneous frequency will vary with time due to the modulation, but the midband frequency f_{c} will remain constant.