10

I will try to give a relatively simple answer to a complex question. I will not give any exact expressions for the error rates for two reasons. The first I stated in my initial sentence, the second is that from your formulas I can see that you only looked for (or at least found) approximate expressions. Judging whether an expression for the error rate is ...


7

The OP's opening statement is incorrect: $f_s > f_{max}/2$ prevents frequency aliasing for a bandlimited signal, but not amplitude aliasing $f_s > 2 f_{max}$ prevents aliasing. It's as simple as that. There is no such distinction as "amplitude aliasing". Since the OP has stated the signal is band-limited; as long as we can assume that means ...


5

If you did a continuous on off keying of a 10101010... pattern, then you would see sidebands as described since this is simply an up-conversion of the Fourier Transform of a 50% duty cycle square wave (moved to any carrier frequency). However if the data pattern for this case of a rectangular on-off keying was random, the resulting spectrum would be ...


5

Amplitude modulation is a linear operation (excluding the carrier in the classical AM) and does not introduce harmonics due to the modulated signal. Sidebands are not harmonics. They are created as the message signal spectrum is shifted up to carrier frequency. Frequency mixing of a message signal with a sinusoidal carrier, another definition of AM, does not ...


5

It's a simple trigonometric identity: $$\cos(\omega_mt)\cos(\omega_ct)=\frac12\left[\cos((\omega_m+\omega_c)t)+\cos((\omega_m-\omega_c))\right]\tag{1}$$ Multiplying two sinusoids of different frequencies results in the sum of two sinusoids with the sum and the difference of the two frequencies. So this type of AM (DSB-SC) really results in a suppressed ...


4

The claim is wrong. Sampling of a pure sinusodial whose frequency is below but arbitrarily close to the Nyquist frequency (half the sampling frequency) is a perfectly valid operation, as long as you can create ideal (zero width transition band) brickwall lowpass filters to be used at the reconstruction interpolation of the continuous waveform from its ...


3

There two definitions for the AM modulated signals The first one is called as the classical-AM (or conventional AM) and is given by $$x_{AM}(t) = (A_c + m(t)) \cdot \cos(2\pi f_c t) \tag{1}$$ and the second one is called as the DSB-SC (double side-band suppressed carrier) and is given by: $$x_{AM}(t) = m(t) \cdot \cos(2\pi f_c t) \tag{2}$$ In your code, you ...


3

$\DeclareMathOperator{\sgn}{sgn}$ The modulating signal in AM is $$s(t) = C + a(t)\text,$$ where $a(t)$ is the (audio) amplitude, and $C$ is a constant so that $s(t) \ge 0 \;\forall t$. (Otherwise, your audio amplitude would just frequently "switch" the wave's sign, not really modulate the envelope.) That means, $C > - \min_t(s(t))$. Therefore, the ...


3

QAM is a digital modulation scheme. As such it is one way of implementing a physical layer that allows to convey digital information over a given medium. QAM is frequently used in all kinds of systems, including wireless (cf. broadcast TV and yes, also WiFi) as well as wired (Ethernet uses some variations of QAM as well). What kind of information you convey ...


2

Because of the diode at the input, you get a DC offset at the output. Notice that a single diode is already a primitive rectifier, because it blocks the negative half wave. But you usually will want a purely AC output, that's why you put a HPF with very low cutoff frequency after the LPF. It will block the DC and yield a pure AC output signal. You can see ...


2

In the frequency domain, an AM modulated sinusoid looks like a carrier plus two side-bands in complex conjugate symmetry above and below the carrier. If you permit the lower sideband to cross into the negative frequency spectrum, then there is no limit to maximum modulating frequency. Of course that may not look like your typical AM modulated signal in the ...


2

The total power of an amplitude modulated signal is $$\begin{align}\overline{s^2_{AM}(t)}&=\overline{\big(A+m(t)\big)^2\cos^2(2\pi f_ct)}\\&=\frac12\overline{\big(A+m(t)\big)^2}+\frac12\overline{\big(A+m(t)\big)^2\cos(4\pi f_ct)}\tag{1}\end{align}$$ The second term on the right-hand side of $(1)$ is zero if $m(t)$ is a lowpass signal, and if $f_c$ is ...


1

Well, you can of course instead of using the Discrete Fourier Transform as filter bank just employ bandpass filters to extract your individual tones. (The DFT is really just a filter bank, if you think about it: it correlates an input sample vector with different complex oscillations. Which is convolution with their conjugate time inverse, which is a filter ...


1

I'll give this a shot. The Fourier Transform of a Gaussian is also a Gaussian. The standard deviations in each domain are related as $\sigma_t \cdot \sigma_F = \frac{1}{2\pi}$ The time standard deviation, $\sigma_t$ has units of time and the frequency domain standard deviation $\sigma_F$ has units of Hz. We can define the "bandwidth" of a gaussion ...


1

The fact that it's modulated with a sinusoid doesn't change the FWHM bandwidth of your pulse – the $e^{jx}$ function has $\left\lvert e^{jx}\right\rvert\equiv 1$ at every point. That doesn't change the amplitude, so the FWHM of a sinusoid-modulated gaussian is just the same as of the unmodulated gaussian.


1

Time-domain multiplication of signals that are each a sum of only a few sinusoidal components is simple to understand as frequency-domain convolution: showing first your signal 1: $$\sin(2\pi f_m t) \cdot \sin(2\pi f_c t),$$ and then your signal 2: $$\big(1-\sin(2\pi f_m t)\big) \cdot \sin(2\pi f_c t),$$ noting that $\cos(x) = \frac{1}{2}e^{-ix} + \frac{1}{...


1

For small angles (see further explanation at end for further details on small angle approximation) the sidebands for phase modulation are closely related to the sidebands for amplitude modulation as revealed in the IQ phasor diagrams below. Both diagrams show large carrier AM and PM modulation being modulated by a single sinusoidal tone, resulting in two ...


1

If I understand correctly, you want to use each side band as an independent information-bearing signal. In other words, you want to be able to generate $$a_k e^{2\pi f_i t}$$ with the information in carried by the amplitude $a_k$. There are several ways to do this, but I don't know which are feasible using optical processing. All of the following methods are ...


1

The claim is generally false. This is studied in details in a 1997 paper by B. Picinbono: On instantaneous amplitude and phase of signals Let $m(t)$ be a positive function corresponding to the information o be transmitted. By multiplying the carrier frequency signal $cos(\omega_0 t)$ by $m(t)$, we obtain the signal $x(t) = m(t) > cos(\omega_0 t)$ ...


1

Make sure that you understand the conditions under which $$\mathcal{H}\big\{A(t)\cos(\omega_0 t)\big\}=A(t)\mathcal{H}\big\{\cos(\omega_0 t)\big\}=A(t)\sin(\omega_0t)\tag{1}$$ holds. Eq. $(1)$ holds if $A(t)$ is a low-pass signal with a cut-off frequency smaller than $\omega_0$. This implies that $A(t)\cos(\omega_0t)$ is a band-pass signal with no energy ...


1

after a RF down-conversion using heterodyning principle, will there be an information loss? No, if the original signal was band-limited, and the bandwidth of your IF processing is sufficiently large to capture that. (so, if you've built a sensible heterodyne receiver.) If we translate to a fixed IF frequency, is there is an intuitive way of explain that ...


1

You need to distinguish between: The error rate for the inner symbols - the error is half of the overlapped segments between the symbol to its closest neighbors (by symmetry we consider one and multiply by 2). $$P_{e1} = Pr(|n|>100-G) = 2*Pr(n>100-G) = 2 \frac{100 - G}{200}$$ The error rate for the points 31G and -31G - same as above bu only for one ...


1

I am not sure if you can use $P_{err}=\text{erfc}(G/\sigma)$ because noise is not gaussian distributed. Here is my take on it. Assuming uniform probability of transmission for all 32 symbols $x_i$,the received signal $y=x_i+n$ so given that $x_i$ was transmitted $y$ is also uniformly distributed in the interval $[-100+x_i,100+x_i]$. Suppose say the ...


1

Define this Tn=0:Ts:1; to be Tn=0:Ts:1 -Ts; By defining it the way you have in the program you have not included an integer number of cycles in the samples. This will result in spectral leakage because your sinuoid now does not fit into one frqeuency bin and thus spill out into other bins and this reduces in magnitude.


1

Yes you are correct that the reception of multiple CDMA signals from different transmitters tends toward a complex Gaussian distribution given the central limit theorem at play - which is not very different from the thermal noise that a spread spectrum signal may be buried in at reception. Yet this does not preclude us from making a hard decision on each ...


1

I see how you added the carrier by multipying by $1 + 0.5cos(2\pi f t)$ (or you used sine, wouldn't change it), and that approach seems fine to me. It looks like you do actually see the carrier in your plot! What I see from your plot does appear to be a signal at +/-25 Hz which is what we would expect to see for $cos(2\pi 25 t)$ (or sine if you used that) ...


1

@A Q. To make your life easier, I suggest you stop using the cosd() command and only use MATLAB's cos() command. For your AM code, if variable dt1 is measured in seconds then your Fs sampling rate is 10,000 samples per second. But your carrier frequency fc1 is set to 500,000. If your fc1 = 500,000 is measured in Hz (cycles/second) then your fc1 value ...


1

The difference is the expected input. The cosd function expects the input to be expressed in degrees, and the cos function expects the input to be in radians. So you have cosd(theta)==cos(deg2rad(theta)). This is just from the MATLAB documentation page. Are you using these correctly? For example, you do cosd(2*pi*F*t). So we have $\frac{2\pi \text{ radians}}...


Only top voted, non community-wiki answers of a minimum length are eligible