# Tag Info

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If you really want to implement digital modulation schemes yourself, you could do it like that: 1) Map a bitstream to a symbol stream. Let's say your bits are A[k] $\in \{0,1\}$. Let's further say you're trying to represent a modulation scheme with d bits per symbol. Then the first step is to map adjacent strings of d bits to numbers $[1,2^d]$. This is done ...

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The QPSK implementation, so you're saying that all your $A[k] = \exp(i \frac{\pi}{4})$ as initialization then you go ahead and multiply with $\exp(i \frac{\pi}{2})$. This is not correct. Let $A$ be a stream of bits then the following loop (the following vector contains 100 bits) A = randi([0 1],1,100) Then a QPSK could be computed in a very similar manner ...

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To complement the other quite appropriate answers: Say you build a real signal $s(t)$ with bandwidth $B$ that transmits $R$ bits per second. Its spectrum $S(f)$ extends from $-B$ to $B$. The modulation property of the Fourier Transform tells you that the spectrum of $$s(t)\cos(2\pi f_c t)$$ is $$\frac{1}{2} \large[ S(f+f_c)-S(f-f_c) \large].$$ Note that ...

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Because bit rate depends on the bandwidth and not on the carrier frequency. Of course at higher frequencies you have more bandwidth, and thus you can transmit more data. But 1 MHz in lower frequencies and 1 MHz at higher frequencies have no difference on the data rate. Other effects may need to be taken into consideration though at higher frequencies. For ...

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If you are a bass and sing A 220 Hz and B 247 Hz at 120 beats per minute, you can sing data at a certain rate. If you are a soprano and sing two quarter notes A 880 Hz and B 988 Hz at the same BPM, your data rate isn't any higher. If you sing more notes (higher bandwidth) you could communicate a more complex score (e.g. a higher data rate). So bandwidth (...

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Use time-domain modulation to shift the frequency before the FFT: Multiply the time domain signal with a complex sinusoid having a period of 2x the FFT frame duration. This will shift the signal cyclically in the frequency domain by 1/2 of the FFT spacing. Carriers on the far edges of the spectrum may be corrupted, but in LTE those are padding carriers. ...

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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 ...

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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 ...

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Its IES or Gate question having last option which you didn't mention here as "carrier frequency can not disappear." . Here correct answer among these 4 option is b i.e. the amplitude of any sideband depends on modulation index . . 2.Frequency spectrum of FM is analysed using Bessel's function .FM spectrum theoretically has infinite number of sidebands . Each ...

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