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Suppose x[n] = cos(2*pi*f0*n/fs) where f0 is the frequency of your sinusoid in Hertz, n=0:N-1, and fs is the sampling rate of x in samples per second. Let X = fft(x). Both x and X have length N. Suppose X has two peaks at n0 and N-n0. Then the sinusoid frequency is f0 = fs*n0/N Hertz.
5 Μαρ 2016 · What you need to know is that the signal $$s(t)=e^{j2\pi f t}$$ has frequency $f$. At any time $t_0$, $s(t_0$) is a point on the unit circle. As time advances, this point moves on the circle, and it completes $f$ turns in one second. Knowing this, you can easily see that the maximum frequency in your signal $x_c(t)$ is 9000 Hz.
3 Φεβ 2015 · To find the max freq, calculate the longest output path delay. Here it is: Tclk2q + Txor + Tsu - Tskew = 4 + 2 + 2 - 1 = 7ns. Max freq of operation = 1/7ns = 142.857 MHz
(a) If X(w) = 0 for Io > W, find the maximum value of T, We, and A such that x,(t) = x(t). (b) Let X 1(w) = 0 for I > 2W and X 2(w) = 0 for Iw > W. Repeat part (a) for the following. (i) x(t) = x1(t) * x 2(t) (ii) x(t) = x1(t) + x 2(t) (iii) x(t) = x 1(t)x 2(t) (iv) x(t) = x 1(lOt)
The relationship of the speed of sound \(v_w\), its frequency \(f\), and its wavelength \(\lambda\) is given by \(v_w = f\lambda,\) which is the same relationship given for all waves. In air, the speed of sound is related to air temperature \(T\) by \(v_w = (331 \, m/s) \sqrt{\dfrac{T}{273 \, K}}.\) \(v_w\) is the same for all frequencies and ...
The maximum frequency in h (t) is , therefore we have the Sampling Theorem: The sampling frequency must be greater than twice the highest frequency in the input waveform in order for all frequencies in the input to be correctly resolved.
9 Οκτ 2024 · The frequency calculator will let you find a wave's frequency given its period or its wavelength and velocity in no time. You can choose a wave velocity from the preset list, so you don't have to remember.
