I've never been explicitly told what the conjugation of a signal in the time-domain means. I'm mainly asking because in my signals class, my professor stated that for a signal x(t) to be real: x(t) = x(t)*, and I don't understand why.
[Math] What does conjugation in the time-domain of a signal mean
educationsignal processing
Related Solutions
The insight and engineering rule of thumb is that
we need to observe at least half of a sine wave to determine its amplitude and frequency.
Of course, as a rule of thumb it is a drastic cutoff, but it is quite intuitive and close to "real".
Now take the sum of two sinusoids with near frequencies
$$
\sin \left( {2\pi \left( {f - \Delta f/2} \right)t} \right) + \sin \left( {2\pi \left( {f + \Delta f/2} \right)t} \right) = 2\cos \left( {2\pi \,\Delta f/2\,t} \right)\sin \left( {2\pi \,f\,t} \right)
$$
we obtain a sine of frequency $f$ amplitude-modulated by a (co)sine of frequency $\Delta f/2$.
(a well known phenomenon of which I do not remember the name in english).
According to the given rule, we need to "see" at least half of the modulating wave to be able and reconstruct the signal, and thus the difference in frequency. If the duration , i.e. "persistance" of the signals, is much less we will only see one sinusoid with the average frequency and double amplitude.
Calling $T$ the duration of the observation window we shall have $$ {1 \over 2}{1 \over {\Delta f/2}} < T\quad \Rightarrow \quad 1 < T\Delta f $$
Passing to a more rigorous mathematical analysis, consider that the Box function (a window of duration $T$) $$ R(t,T) = \left\{ {\matrix{ 1 & { - T/2 \le t \le T/2} \cr 0 & {otherwise} \cr } } \right. = U(t + T/2) - U(t - T/2) $$ where $U$ is step function, has a frequency spectrum (bilateral Fourier Transform) given by $$ G(f,T) = T{\rm sinc}(f\,T) = {{{\rm sin}(\pi f\,T)} \over {\pi f}} $$
A sinusoid of duration $T$ is the product of the Box function for
a sine function.
The spectrum will be the convolution of a Dirac at frequency $f$ (apart that at $-f$)
with the $G(f,T) =T{\rm sinc}(f\,T) $ function , i.e. the same centered at $f$.
Then, approximating the $G(f,T)$ to a box function with a cutoff at $f \pm 1/(2T)$ leads to that we can distinguish between frequencies with a separation of at least $1/T$, i.e. to the formula given above.
After some work and help from a colleague, I managed to get an answer. For posterity, here it is:
And apparently, yes, we're supposed to use convolution from the fact that we're asked for impulse response.
% Creating our sine-signal
clear
t = -50:50;
Fs = 25;
x=sin(2*pi*(1/Fs)*t);
% Creating our impulse responses. They're unit steps, going from all-zeroes to all-ones.
h1 = [zeros(1,50) ones(1,50)];
h2 = [zeros(1,40) ones(1,60)];
% Calculating the convolusion result of the square signals above with our original sine-signal.
convPlot1 = conv(x,h1);
convPlot2 = conv(x,h2);
% Plot and check our result.
plot(convPlot1)
plot(convPlot2)
Best Answer
If $$ z\equiv a+ib $$ is a complex number ($a$ and $b$ are real numbers), its conjugate is defined to be $$ \overline{z}\equiv a-ib. $$
Note that we can write a real number $x$ as $$ x=x+i0. $$ The conjugate of $x$ is equal to itself: $$ \overline{x}=x-i0=x. $$ Therefore, if a number is real, it is equal to its conjugate.
For the converse, if $z\equiv a+ib$ is equal to its conjugate, $$ a+ib=a-ib $$ which implies $$ ib=-ib. $$ If $b$ is anything other than zero, we arrive at a contradiction, since $i\neq-i$. Therefore, if a complex number is equal to its conjugate, it is real.
All that remains to be done is to apply your result pointwise (i.e. for every possible time $t$), and you arrive at the result $x(t)=x(t)^*$ for all $t$ if and only if $x(t)$ is real for all $t$.