## Archive for March, 2011

### Interpolation and energy amplification

March 25, 2011

Consider the following interpolator [1]

where $x_u = (\uparrow L) (x)$ is the upsampling of sequence $x$ by an integer factor of $L$, and where $H(z)$ is the transfer function of an ideal discrete-time LTI lowpass filter with gain $L$ and cutoff frequency $\pi/L$. What is the relation between the energies of signals $x$ and $y$?

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My solution:

It can be easily shown [1] that the interpolator’s input-output relationship is $Y (z) = H(z) X (z^L)$. Hence, evaluating the Z-transforms on the unit circle, we obtain

$Y (e^{j \omega}) = H(e^{j \omega}) X (e^{j \omega L})$.

The energy of the interpolator’s output signal is

$\mathcal{E}_y := \displaystyle \sum_{n \in \mathbb{Z}} | y(n) |^2 = \frac{1}{2 \pi} \int_{-\pi}^{\pi} |Y (e^{j \omega})|^2 d \omega$

where we used Parseval’s theorem for discrete-time signals. Thus,

$\mathcal{E}_y = \displaystyle\frac{1}{2 \pi} \int_{-\pi}^{\pi} |H(e^{j \omega})|^2 |X (e^{j \omega L})|^2 d \omega$.

$H(z)$ is an ideal lowpass filter with the following frequency response

$H(e^{j \omega}) = \begin{cases} L, & |\omega| \leq \pi/L\\ 0, & \pi/ L< |\omega| \leq \pi\end{cases}$

and the following sinc impulse response

$h(n) = \displaystyle\frac{1}{2 \pi} \int_{-\pi}^{\pi} H(e^{j \omega}) e^{j \omega n} d \omega = \displaystyle \frac{\sin(\frac{n \pi}{L})}{\frac{n \pi}{L}} =: \text{sinc} \left(\frac{n \pi}{L}\right)$.

Please do note that this is a non-causal filter with an impulse response of infinite length. We can then write the interpolator’s output signal’s energy as follows

$\mathcal{E}_y = \displaystyle\frac{1}{2 \pi} \int_{-\pi}^{\pi} |H(e^{j \omega})|^2 |X (e^{j \omega L})|^2 d \omega = \displaystyle\frac{L^2}{2 \pi} \int_{-\pi/L}^{\pi/L} |X (e^{j \omega L})|^2 d \omega$.

Performing a change of variable, $\theta = L \omega$, we obtain

$\mathcal{E}_y = \displaystyle\frac{L^2}{2 \pi} \int_{-\pi/L}^{\pi/L} |X (e^{j \omega L})|^2 d \omega = \displaystyle\frac{L}{2 \pi} \int_{-\pi}^{\pi} |X (e^{j \theta})|^2 d \theta = L \mathcal{E}_x$

where

$\mathcal{E}_x := \displaystyle \sum_{n \in \mathbb{Z}} | x(n) |^2 = \frac{1}{2 \pi} \int_{-\pi}^{\pi} |X (e^{j \omega})|^2 d \omega$

is the energy of the input signal. Since $\mathcal{E}_y = L \mathcal{E}_x$, we can conclude that the interpolator amplifies the input signal’s energy by a factor of $L$, which is somewhat intuitive.

We considered the case where signals are bi-infinite sequences and $H(z)$ is a non-causal ideal lowpass filter. It would be interesting to consider the more realistic case where the signals are finite sequences and $H(z)$ is a causal FIR filter.

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References

[1] Alan V. Oppenheim, Ronald W. Schafer, Discrete-Time Signal Processing, 2nd edition, Prentice-Hall, 1999.

### Oddly real

March 12, 2011

Consider a complex-valued continuous-time signal, $x : \mathbb{R} \to \mathbb{C}$, with the property $x(t) = - x^*(-t)$. Show that the even part, $x_e$, and the odd part, $x_o$, of signal $x$ are imaginary and real, respectively. Recall that the even and imaginary parts of a signal are defined as follows [1]

$x_e (t) := \displaystyle\frac{x(t) + x(-t)}{2}, \qquad{} x_o (t) := \displaystyle\frac{x(t) - x(-t)}{2}.$

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My solution:

Since $x$ is a complex-valued signal, we can decompose it into its real and imaginary parts

$x(t) = x_R (t) + j x_I (t)$

where $x_R$ and $x_I$ are real-valued signals. The complex conjugate of $x$ is

$x^*(t) = x_R (t) - j x_I (t)$

and, from $x(t) = - x^*(-t)$, we know that $x(-t) = - x^*(t)$. The even part is, then

$x_e (t) = \displaystyle\frac{x(t) + x(-t)}{2} = \displaystyle\frac{x(t) - x^*(t)}{2} = j x_I (t)$

which is purely imaginary, and the odd part is

$x_o (t) = \displaystyle\frac{x(t) - x(-t)}{2} = \displaystyle\frac{x(t) + x^*(t)}{2} = x_R (t)$

which is real.

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References

[1] Alan V. Oppenheim, Alan S. Willsky, S. Hamid Nawab, Signals & Systems, 2nd edition, Prentice-Hall, 1996.