As the DFT of a signal amounts to being a weighted sum, it follows naturally that the DFT operation is linear. So if we have two arbitrary signals $x_1[n]$ and $x_2[n]$ (with DFTs $X_1[k]$ and $X_2[k]$) and arbitrary constants $\alpha_1$ and $\alpha_2$, then the DFT of $\alpha_1 x_1[n]+\alpha_2 x_2[n]$ is simply $\alpha_1 X_1[k]+\alpha_2 X_2[k]$:
$x_1[n]~\stackrel{\textrm{DFT}}{\longleftrightarrow}~ X_1[k], ~~x_2[n] ~\stackrel{\textrm{DFT}}{\longleftrightarrow}~ X_2[k]\\
\alpha_1 x_1[n]+ \alpha_2 x[2] ~\stackrel{\textrm{DFT}}{\longleftrightarrow}~ \alpha_1 X_1[k]+ \alpha_2 X_2[k]$The proof is straightforward:
$\begin{align*}\textrm{DFT}\{\alpha_1 x_1[n]+ \alpha_2 x_2[n]\}&=\sum_{n=0}^{N-1} (\alpha_1 x_1[n] + \alpha_2 x[2]) e^{-j \frac{2\pi}{N}kn}\\&=(\alpha_1\sum_{n=0}^{N-1} x_1[n]e^{-j \frac{2\pi}{N}kn}) + (\alpha_2\sum_{n=0}^{N-1}x_2[n]e^{-j \frac{2\pi}{N}kn})\\&=\alpha_1 X_1[k] + \alpha_2 X_2[k]\end{align*}$
The convolution/multiplication time/frequency relationship
We now reach perhaps the most significant DFT property, the relationship between convolution and multiplication in time and frequency. Suppose we have 2 discrete-time finite length signals $x_1[n]$ and $x_2[n]$, whose DFTs are $X_1[k]$ and $X_2[k]$. Let $y[n]$ be the circular convolution $y[n]=x_1[n]\circledast x_2[n]$. Then the DFT of $y[n]$ is equivalent to the product of the DFTs of $x_1[n]$ and $x_2[n]$: $Y[k]=X_1[k]X_2[k]$. Or, to express this relationship in DFT pair notation, we have:
$x_1[n]\circledast x_2[n] \stackrel{\textrm{DFT}}{\longleftrightarrow} X_1[k]X_2[k]$The proof of this relationship will use the $r=(n-m)_N=n-m+lN$ change of variable we used earlier.
$\begin{align*}\textrm{DFT}\{x_1[n]\circledast x_2[n]\}&=~ \sum_{n=0}^{N-1} \left( \sum_{m=0}^{N-1} \: x_1[(n-m)_N] \, x_2[m]\right) e^{-j\frac{2\pi}{N}k n} \\&= \sum_{m=0}^{N-1} x_2[m] \left( \sum_{n=0}^{N-1} x_1[(n-m)_N]e^{-j\frac{2\pi}{N}k n} \right)\\&=\sum_{m=0}^{N-1} x_2[m] \left( \sum_{r=0}^{N-1} x_1[r]e^{-j\frac{2\pi}{N}k (r+m+lN)} \right) \\&=\sum_{m=0}^{N-1} x_2[m] \left( \sum_{r=0}^{N-1} x_1[r]e^{-j\frac{2\pi}{N}kr}e^{-j\frac{2\pi}{N}k m}e^{-j\frac{2\pi}{N}klN} \right) \\&= \left( \sum_{m=0}^{N-1} x_2[m] e^{-j\frac{2\pi}{N}km} \right) \left( \sum_{r=0}^{N-1} x_1[r]e^{-j\frac{2\pi}{N}kr} \right)\\&=X_1[k]X_2[k]\end{align*}$
So we see that convolution in two signals' time corresponds to their multiplication in frequency. It is certainly an interesting relationship, but it is also very practical: $N$ multiplications in the frequency domain is obviously much than the $N^2$ multiplications required to compute for an $N$-length signal. Of course, this requires first that the DFT of the two signals be computed, but we will see there are ways to compute that efficiently. What it all means is that a system's output (which is found via convolution) can be computed efficiently by way of multiplication in the frequency domain.
Dft symmetry properties
As the DFT is a weighted sum of complex signals (complex harmonic sinusoids), it follows that the DFT of signals is, in general, complex-valued.
A real-valued finite-length signal $x[n]$.CAPTION.CAPTION.CAPTION. Perhaps you have noticed, in the figures above, a certain symmetry for the signal's DFT. For the real part of the DFT, you see that $Re[X[k]]=Re[X[N-k]]$, and that $Im[X[k]]=-Im[X[N-k]]$. Recalling that the DFT is periodic, if we consider frequencies from $-\frac{N}{2}$ to $\frac{N}{2}$, then we have $Re[X[k]]=Re[X[-k]]$, and that $Im[X[k]]=-Im[X[-k]]$. Or, in words, the real part of the DFT is even, and the imaginary part is odd.
A real-valued finite-length signal $x[n]$.CAPTION.CAPTION.CAPTION.
This symmetry was not a coincidence for the DFT of that particular $x[n]$; because the complex harmonic sinusoid signals that make up the DFT sum formula are conjugate symmetric (the real part is even, the imaginary part is odd), the DFTs of all purely real signals will also be conjugate symmetric. The converse is true for purely imaginary signals; their DFTs will be symmetric such that the real part is odd and the imaginary part is even. If, in addition to being purely real or purely imaginary, the signal itself is also even or odd, that will further limit the characteristics of its DFT. All of these DFT symmetry properties are listed on the table below:DFT SYMMETRY TABLE
Questions & Answers
A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
