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The spectrum of a continuous tone is, as you said, of the form $\delta(f-f_0) + \delta(f+f_0)$: 2 impulses at frequencies $f_0$ and $-f_0$. As a lowpass signal, this is said to have bandwidth $f_0$ (the one-sided spectrum has components up to $f_0$). As a bandpass signal, it has zero bandwidth (there's nothing around the carrier frequency $f_0$). If you ...
That actually would be a great interview questions. Let's derive it from scratch: $\text{sinc}(x)$: The Fourier Transform of a $\text{sinc}(x)$ is a rectangle. The normalized $\text{sinc}()$ function results in a bandwidth of 1 Hz (from -0.5Hz to 0.5Hz) $t-10$: The $(t-10)$ term is irrelevant to the bandwidth: time shift is equivalent to multiplying with $e^... 12 The bandwidths simply add. You can break down both signal into their sinusoidal components and pairwise multiply them. For every pair of sines you get the sum and difference frequencies. This is a simple consequence of the multiplication theorems for sine waves. The highest frequency of the product will be the sum of the highest frequencies in the ... 9 Remember that Wavelet Transforms are nothing but time-localized filtering/correlation operations. The wavelet transforms provide a unified framework for getting around the Heisenberg Uncertainly Principle that the Fourier Transform suffers from. So when you ask "what should my settings be for bandwidth, and center frequency", you are asking for filter ... 9 A sinusoidal signal is represented as $$x(t) = \mathrm{cos}(\omega t) = \mathrm{cos}(2\pi f t)$$$\omega$is the angular frequency and$fis the frequency. See Frequency definition. Your signal \begin{align} x(t) &= 5 + 30\mathrm{cos}(2000\pi t) + 10\mathrm{cos}(6000\pi t)\\ &= 5\mathrm{cos}(2\pi \times 0 t) + 30\mathrm{cos}(2\pi \times 1000 t) + ... 8 The bandwidth of a theoretical infinite length sinusoid of a perfectly constant frequency is zero. The bandwidth of a time-limited sinusoidal pulse is the transform of the pulse envelope. For a rectangular time window, that transform is a Sinc function. The main lobe of that Sinc is about 2/t in bandwidth, but that contains only a portion of that Sinc's ... 8 An ideal sinusoid does indeed have zero bandwidth, but it is not useful to transmit one because it also imparts zero information to the receiver. Once the receiver knows what the sinusoid phase is it can perfectly predict the rest of the signal- thus, no more information is received. For information to be transmitted there has to be an element of "not ... 7 I will try to derive your formula. The formula you were given is technically not correct, but on a practical level it is. I will try to explain below. The formula for the frequency of any wave (sound, water, light, etc.) isf = \frac{\nu}{\lambda}$, where$\lambda$is the wavelength and$\nu$is the wave's velocity. This makes intuitive sense if you ... 7 The$BT$product is the bandwidth-symbol time product where$B$is the$-3\textrm{ dB}$(half-power) bandwidth of the pulse/filter and$T$is the symbol duration. For different applications you will find varying recommended values. In GSM telephony for instance, a$BT=0.3$is recommended. In satellite communications with GMSK, for near-earth missions the ... 6 A recording originally at 8kHz and digitally upsampled to 16kHz will have almost no energy in the 4-8kHz range (whatever is here is due to imperfections in the filters used for the upsampling process). I would just use a 4kHz and 5.5kHz high pass; and use a threshold on the signal energy at the output of these filters. ... Unless your recordings are ... 6 It's generally not true that a band pass signal can be sampled and recovered without error if$f_s>2B$is satisfied, where$B=f_h-f_lis the signal's bandwidth. This condition is just necessary but not sufficient. You have to make sure that the aliased spectra do not overlap. This results in the following condition on the sampling frequency: $$\frac{... 6 In the Nyquist sampling theorem, the bandwidth is usually defined as the largest frequency in the signal; in other words, what the sampling theorem says is that, if you sample this signal at 6.8 kilosamples per second, you will be able to reconstruct the analog signal perfectly. Now, under some conditions, you can do bandpass sampling, where the bandwidth ... 6 Short answer 10*log(bw/fs) to take into account the oversampling operation because the awgn() function specifies the signal-to-noise ratio per sample, in dB. Longer answer The discrete time AWGN model is$$Y = X+N$$where X is data from continuous time X(t), N is noise sequence from AWGN process N(t) and Y is receive symbols. If X(t) is ... 5 First a demonstration that the squares of both$$\begin{align}&[\dots, 0, 0, 1,\hphantom{-}1, 0, 0, \dots] \text{ and}\\ &[\dots, 0, 0, 1, -1, 0, 0, \dots]\end{align}$$equal$$[\dots, 0, 0, 1, \hphantom{-}1, 0, 0, \dots]\hphantom{\text{ and}}$$but the squares of their sinc interpolations differ (Fig. 1): Figure 1. Squares of sinc interpolations ... 5 Not at all. The Uncertainty Principle says that a function cannot be both limited in time and limited in frequency. More specifically, the product of the signal's widths in time and in frequency (i.e., its time extension \Delta_t and its bandwidth \Delta_f) is bounded from below:$$\Delta_t\cdot\Delta_f\ge C\tag{1}$$where the constant C depends on ... 5 The final statement is correct, for PSK with proper pulse shaping the baud rate and the bandwidth are the same (the bandwidth will typically be 20-30% higher than the symbol rate, but read on). The baud rate is the symbol rate or the rate of change from one symbol to the next in the modulated constellation. Your earlier figures suggest unfiltered waveforms ... 5 Does this formula mean that these two channels have the same transmission speed? Yes. That's exactly what you should take away from this: Channel capacity has nothing to do with center frequency; only bandwidth and SNR are relevant. Cause I think the one from 1 GHz to 1.02 GHz is faster, since it has a higher frequency. But that's plain wrong. It's not ... 4 Dilip's points in his answer are correct. Speaking more to the context that you referenced of pulse-compression radar, I think you're getting confused by differing meanings of the oft-used word "resolution." In a broad signal processing sense, your time resolution is defined to some extent by your sample rate. But in the specific problem domain of building a ... 4 In order to avoid inter symbol interference (ISI) the baseband transmit signal must meet Nyquist's first criterion. In frequency domain it can be formulated as follows.$$ \omega_\mathrm{N} = \frac{v_\mathrm{s}}{2} With v_\mathrm{s} the symbol rate and \omega_\mathrm{N} the 3dB frequency of a low pass filter with symmetric flank. The flank must be ... 4 Unless I'm missing something here it would just be the difference in frequency between the highest frequency sine wave and the lowest: BW = Fmax - Fmin 4 The need for modulation: Your voice signals generally lies in the frequency range 1kHz to 4kHz and music lies in the range 20Hz to 20Khz. Say your locality has 4-5 AM broadcast stations. Now all the AM broadcast stations cannot transmit their content as is because if all the stations transmit at the same time there will be signal interference. Also, for ... 4 In communications, whether a signal is narrowband (has a small bandwidth) or wideband (has a large bandwidth) is relative to the channel's impulse response. If the channel behaves as a flat filter (constant gain) over the signal's bandwidth, then the signal is narrowband. If the channel introduces significant distortion (that is, if different signal ... 4 Do you perhaps mean small signal model? That is often used in the context of analysing the performance of transistor circuits. The transistors themselves are nonlinear devices, but they can be analysed as linear devices if you choose their operating regime appropriately. Conversely, the large signal model is when the operating regime is such that the ... 4 Interpolated points of the DFT can be computed using a dot-product of a few samples around the peak region with a pre-computed interpolation vector. The interpolation vector is determined by the location of the desired interpolated sample, taking into consideration the amount of zero-padding required, etc., etc. This technique and the methodology to ... 4 To complement my part to this question: Here is a somewhat shorted answer based upon a manual expansion of the odd function f(x) \begin{align*} f(x)&=\ln\left(\arctan\left(\alpha e^x\right)\right)-\ln\left(\arctan\left(\alpha e^{-x}\right)\right)\\ &=f_1x+f_3x^3+O\left(x^5\right)\tag{1} \end{align*} into a series up to the third order. Some ... 4 A theorem, which I know as Weyl's, 1931, is: if s(t) and derived functions ts(t), s'(t) are in L^2 with the related \|\cdot\| norm symbol then: \| s(t) \|^2 \leq 2\| ts(t) \| \| s'(t) \|$$Equality is attained when s(t) is a modulated Gaussian/Gabor elementary function defined as:$$ s'(t) / s(t) \propto t $$or practically as:$$s(t) = ... 4 This picture may help clear up the OP's confusion. On the right is the baseband signal of interest, with finite bandwidth centered at DC as shown in the frequency spectrum plot. This is a complex signal, so it has I (in-phase, on the real axis) and Q (quadrature, on the imaginary axis) components. We implement these signals in hardware using two real ... 4 You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92). With a well-behaved signalf(t)$as you defined it above (in particular,$f(t)$is integrable and bandlimited to$B\,\text{Hz}$, and$\text{sup}\,|f(t)| = A$), then $$\left|\frac{\text{d}f(t)}{\text{d}t}\right| \leq 2AB\... 4 Yes, of course, you can sum them. The bandwidth of the resulting signal is simply the min/max of the individual signals. If we assume$$z(t)=x(t)+y(t)$$Then then bandwidth of$z(t)$will simply be$[min(f_{min},g_{min}),max(f_{max},g_{max})]$, so in general the you will have$B_z > B_x$Keep in mind that for any real valued signal$x(t)$the spectrum ... 3 Find the Fourier Transform$F(\omega)$and perform the following integration.$\frac{1}{2\pi}\int_{-W}^{W}|F(\omega)|^2\mathrm{d}\omega=0.7E$, where$E$is the total Energy of the function given by:$E=\frac{1}{2\pi}\int_{-\infty}^{+\infty}|F(\omega)|^{2}\mathrm{d}\omega$.The value of$W\$ is the 70% Bandwidth.