Solving DSP problem
in [DSP]
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From: Tim Wescott on 19 Jan 2010 12:15 On Mon, 18 Jan 2010 23:06:30 -0800, Andor wrote: > On 18 Jan., 22:58, Tim Wescott <t...(a)seemywebsite.com> wrote: >> On Mon, 18 Jan 2010 15:51:21 -0500, Jerry Avins wrote: >> > Tim Wescott wrote: >> >> On Mon, 18 Jan 2010 13:40:37 -0500, Jerry Avins wrote: >> >> >>> Tim Wescott wrote: >> >>>> On Mon, 18 Jan 2010 09:25:50 -0600, Richello wrote: >> >> >>>>> Dear Sir, >> >> >>>>> Could you please help me to solve this or give me hints to do so? >> >> >>>>> An analogue signal x(t)=10cos(500πt) is sampled at 0, T,2T, .... >> >>>>> with T=1ms. >> >> >>>>> I want to find a cosine y(t), whose frequency is as close as >> >>>>> possible to that of x(t), which when sampled with T=1ms yields >> >>>>> the same sample values as x(t). how can I get the equation of >> >>>>> y(t)? .. then if x(nT) were the input to a D/A converter, >> >>>>> followed by a low-pass smoothing filter, why would the output be >> >>>>> x(t) and not y(t) ? >> >>>> 1: The solution as you state the problem is trivial: y(t) = x(t). >> >>>> I assume you mean the frequency should be close to but different. >> >> >>>> 2: Have you asked your prof? >> >> >>>> 3: Read this: http://www.wescottdesign.com/articles/Sampling/ >> >>>> sampling.html. Skip down to the part about aliasing. >> >>> First, fix the link: >> >>>http://www.wescottdesign.com/articles/Sampling/sampling.html >> >> >>> I'm puzzled by Tim's reference to aliasing. The frequency is 250 Hz >> >>> and the sample rate is 1000 Hz. >> >> >>> Jerry >> >> >> He's asking for a continuous-time signal (presumably not identical >> >> to the given one) that gives the same discrete-time signal after >> >> sampling. >> >> That sounds like aliasing to me. >> >> >> It sounds like a _homework problem_ about aliasing. One I might >> >> write were I teaching a signal processing course, I might add. >> >> > Got it. I would have worded it differently, though. >> >> I would have worded it differently, too. A set of homework problems >> that doesn't contain at least one veiled reference to Bart Simpson, >> Batman, Diogenes, Gilligan, or some other major philosopher is a minor >> failure, IMHO. > > Cool - can you reword the OPs problem in that vein? I don't know if I can or not, come to think of it. But one of the recurring gags in Gilligan's Island was when Gilligan would come to someplace (a hut, a clearing) and do something important -- repeatedly -- when no one was there to see. In between these events, someone else (usually the Skipper) would come to the clearing and be frustrated at Gilligan's absence. It's a classic case of aliasing, or at least under sampling. -- www.wescottdesign.com
From: Tim Wescott on 19 Jan 2010 13:45 On Mon, 18 Jan 2010 23:06:30 -0800, Andor wrote: > On 18 Jan., 22:58, Tim Wescott <t...(a)seemywebsite.com> wrote: >> On Mon, 18 Jan 2010 15:51:21 -0500, Jerry Avins wrote: >> > Tim Wescott wrote: >> >> On Mon, 18 Jan 2010 13:40:37 -0500, Jerry Avins wrote: >> >> >>> Tim Wescott wrote: >> >>>> On Mon, 18 Jan 2010 09:25:50 -0600, Richello wrote: >> >> >>>>> Dear Sir, >> >> >>>>> Could you please help me to solve this or give me hints to do so? >> >> >>>>> An analogue signal x(t)=10cos(500πt) is sampled at 0, T,2T, .... >> >>>>> with T=1ms. >> >> >>>>> I want to find a cosine y(t), whose frequency is as close as >> >>>>> possible to that of x(t), which when sampled with T=1ms yields >> >>>>> the same sample values as x(t). how can I get the equation of >> >>>>> y(t)? .. then if x(nT) were the input to a D/A converter, >> >>>>> followed by a low-pass smoothing filter, why would the output be >> >>>>> x(t) and not y(t) ? >> >>>> 1: The solution as you state the problem is trivial: y(t) = x(t). >> >>>> I assume you mean the frequency should be close to but different. >> >> >>>> 2: Have you asked your prof? >> >> >>>> 3: Read this: http://www.wescottdesign.com/articles/Sampling/ >> >>>> sampling.html. Skip down to the part about aliasing. >> >>> First, fix the link: >> >>>http://www.wescottdesign.com/articles/Sampling/sampling.html >> >> >>> I'm puzzled by Tim's reference to aliasing. The frequency is 250 Hz >> >>> and the sample rate is 1000 Hz. >> >> >>> Jerry >> >> >> He's asking for a continuous-time signal (presumably not identical >> >> to the given one) that gives the same discrete-time signal after >> >> sampling. >> >> That sounds like aliasing to me. >> >> >> It sounds like a _homework problem_ about aliasing. One I might >> >> write were I teaching a signal processing course, I might add. >> >> > Got it. I would have worded it differently, though. >> >> I would have worded it differently, too. A set of homework problems >> that doesn't contain at least one veiled reference to Bart Simpson, >> Batman, Diogenes, Gilligan, or some other major philosopher is a minor >> failure, IMHO. > > Cool - can you reword the OPs problem in that vein? I've got it: Gilligan has sprung a net trap set by natives from the next island over; the trap is hung from a tall tree branch, and Gilligan is swinging over a small clearing with a period of oscillation of 5 seconds (0.2Hz). The Skipper has also been captured and blindfolded and is being held nearby. Every two seconds exactly he manages to pull off the blindfold and see Gilligan, but his captors immediately replace the blindfold each time. What is the _closest_ (but not identical) period of oscillation at which Gilligan could be swinging such that the Skipper would see exactly the same period of oscillation given how often he pulls off his blindfold? For extra credit: The Skipper escapes, and runs to the Professor to explain Gilligan's dilemma (including the timing, of course). The Professor (having taken signal processing courses along with everything else), knows that there are two likely periods to the swing, and thus knows the two possible lengths of the rope from which Gilligan is suspended (he ignores the mass of the rope and assumes the tree is infinitely rigid). Assuming that the professor gets his math right, how long are his two calculated rope lengths? -- www.wescottdesign.com
From: maury on 19 Jan 2010 17:14 On Jan 19, 11:15 am, Tim Wescott <t...(a)seemywebsite.com> wrote: > On Mon, 18 Jan 2010 23:06:30 -0800, Andor wrote: > > On 18 Jan., 22:58, Tim Wescott <t...(a)seemywebsite.com> wrote: > >> On Mon, 18 Jan 2010 15:51:21 -0500, Jerry Avins wrote: > >> > Tim Wescott wrote: > >> >> On Mon, 18 Jan 2010 13:40:37 -0500, Jerry Avins wrote: > > >> >>> Tim Wescott wrote: > >> >>>> On Mon, 18 Jan 2010 09:25:50 -0600, Richello wrote: > > >> >>>>> Dear Sir, > > >> >>>>> Could you please help me to solve this or give me hints to do so? > > >> >>>>> An analogue signal x(t)=10cos(500Ït) is sampled at 0, T,2T, .... > >> >>>>> with T=1ms. > > >> >>>>> I want to find a cosine y(t), whose frequency is as close as > >> >>>>> possible to that of x(t), which when sampled with T=1ms yields > >> >>>>> the same sample values as x(t). how can I get the equation of > >> >>>>> y(t)? .. then  if x(nT) were the input to a D/A converter, > >> >>>>> followed by a low-pass smoothing filter, why would the output be > >> >>>>> x(t) and not y(t) ? > >> >>>> 1:  The solution as you state the problem is trivial: y(t) = x(t). > >> >>>>  I assume you mean the frequency should be close to but different. > > >> >>>> 2:  Have you asked your prof? > > >> >>>> 3:  Read this:  http://www.wescottdesign.com/articles/Sampling/ > >> >>>> sampling.html.  Skip down to the part about aliasing. > >> >>> First, fix the link: > >> >>>http://www.wescottdesign.com/articles/Sampling/sampling.html > > >> >>> I'm puzzled by Tim's reference to aliasing. The frequency is 250 Hz > >> >>> and the sample rate is 1000 Hz. > > >> >>> Jerry > > >> >> He's asking for a continuous-time signal (presumably not identical > >> >> to the given one) that gives the same discrete-time signal after > >> >> sampling. > >> >>  That sounds like aliasing to me. > > >> >> It sounds like a _homework problem_ about aliasing.  One I might > >> >> write were I teaching a signal processing course, I might add. > > >> > Got it. I would have worded it differently, though. > > >> I would have worded it differently, too.  A set of homework problems > >> that doesn't contain at least one veiled reference to Bart Simpson, > >> Batman, Diogenes, Gilligan, or some other major philosopher is a minor > >> failure, IMHO. > > > Cool - can you reword the OPs problem in that vein? > > I don't know if I can or not, come to think of it. > > But one of the recurring gags in Gilligan's Island was when Gilligan > would come to someplace (a hut, a clearing) and do something important -- > repeatedly -- when no one was there to see.  In between these events, > someone else (usually the Skipper) would come to the clearing and be > frustrated at Gilligan's absence. > > It's a classic case of aliasing, or at least under sampling. > > --www.wescottdesign.com- Hide quoted text - > > - Show quoted text - Tim, At first I thought it was an aliasing problem too. But the OP stated "whose frequency is as close as possible to that of x(t)". As close as possible would be equal, and it would be a phasing problem instead of an aliasing problem. Am I missing something here? Maurice Givens
From: Tim Wescott on 19 Jan 2010 17:39 On Tue, 19 Jan 2010 14:14:13 -0800, maury wrote: > On Jan 19, 11:15 am, Tim Wescott <t...(a)seemywebsite.com> wrote: >> On Mon, 18 Jan 2010 23:06:30 -0800, Andor wrote: >> > On 18 Jan., 22:58, Tim Wescott <t...(a)seemywebsite.com> wrote: >> >> On Mon, 18 Jan 2010 15:51:21 -0500, Jerry Avins wrote: >> >> > Tim Wescott wrote: >> >> >> On Mon, 18 Jan 2010 13:40:37 -0500, Jerry Avins wrote: >> >> >> >>> Tim Wescott wrote: >> >> >>>> On Mon, 18 Jan 2010 09:25:50 -0600, Richello wrote: >> >> >> >>>>> Dear Sir, >> >> >> >>>>> Could you please help me to solve this or give me hints to do >> >> >>>>> so? >> >> >> >>>>> An analogue signal x(t)=10cos(500πt) is sampled at 0, T,2T, >> >> >>>>> .... with T=1ms. >> >> >> >>>>> I want to find a cosine y(t), whose frequency is as close as >> >> >>>>> possible to that of x(t), which when sampled with T=1ms yields >> >> >>>>> the same sample values as x(t). how can I get the equation of >> >> >>>>> y(t)? .. then if x(nT) were the input to a D/A converter, >> >> >>>>> followed by a low-pass smoothing filter, why would the output >> >> >>>>> be x(t) and not y(t) ? >> >> >>>> 1: The solution as you state the problem is trivial: y(t) = >> >> >>>> x(t). >> >> >>>> I assume you mean the frequency should be close to but >> >> >>>> different. >> >> >> >>>> 2: Have you asked your prof? >> >> >> >>>> 3: Read this: http://www.wescottdesign.com/articles/Sampling/ >> >> >>>> sampling.html. Skip down to the part about aliasing. >> >> >>> First, fix the link: >> >> >>>http://www.wescottdesign.com/articles/Sampling/sampling.html >> >> >> >>> I'm puzzled by Tim's reference to aliasing. The frequency is 250 >> >> >>> Hz and the sample rate is 1000 Hz. >> >> >> >>> Jerry >> >> >> >> He's asking for a continuous-time signal (presumably not >> >> >> identical to the given one) that gives the same discrete-time >> >> >> signal after sampling. >> >> >> That sounds like aliasing to me. >> >> >> >> It sounds like a _homework problem_ about aliasing. One I might >> >> >> write were I teaching a signal processing course, I might add. >> >> >> > Got it. I would have worded it differently, though. >> >> >> I would have worded it differently, too. A set of homework problems >> >> that doesn't contain at least one veiled reference to Bart Simpson, >> >> Batman, Diogenes, Gilligan, or some other major philosopher is a >> >> minor failure, IMHO. >> >> > Cool - can you reword the OPs problem in that vein? >> >> I don't know if I can or not, come to think of it. >> >> But one of the recurring gags in Gilligan's Island was when Gilligan >> would come to someplace (a hut, a clearing) and do something important >> -- repeatedly -- when no one was there to see. In between these >> events, someone else (usually the Skipper) would come to the clearing >> and be frustrated at Gilligan's absence. >> >> It's a classic case of aliasing, or at least under sampling. >> >> --www.wescottdesign.com- Hide quoted text - >> >> - Show quoted text - > > Tim, > At first I thought it was an aliasing problem too. But the OP stated > "whose frequency is as close as possible to that of x(t)". As close as > possible would be equal, and it would be a phasing problem instead of an > aliasing problem. Am I missing something here? > > Maurice Givens A strict reading gives that answer, yes. I suspect that he (or his instructor) left the "but not equal" out. -- www.wescottdesign.com
From: Andor on 21 Jan 2010 02:55
On 19 Jan., 19:45, Tim Wescott <t...(a)seemywebsite.com> wrote: > On Mon, 18 Jan 2010 23:06:30 -0800, Andor wrote: > > On 18 Jan., 22:58, Tim Wescott <t...(a)seemywebsite.com> wrote: > >> On Mon, 18 Jan 2010 15:51:21 -0500, Jerry Avins wrote: > >> > Tim Wescott wrote: > >> >> On Mon, 18 Jan 2010 13:40:37 -0500, Jerry Avins wrote: > > >> >>> Tim Wescott wrote: > >> >>>> On Mon, 18 Jan 2010 09:25:50 -0600, Richello wrote: > > >> >>>>> Dear Sir, > > >> >>>>> Could you please help me to solve this or give me hints to do so? > > >> >>>>> An analogue signal x(t)=10cos(500Ït) is sampled at 0, T,2T, .... > >> >>>>> with T=1ms. > > >> >>>>> I want to find a cosine y(t), whose frequency is as close as > >> >>>>> possible to that of x(t), which when sampled with T=1ms yields > >> >>>>> the same sample values as x(t). how can I get the equation of > >> >>>>> y(t)? .. then  if x(nT) were the input to a D/A converter, > >> >>>>> followed by a low-pass smoothing filter, why would the output be > >> >>>>> x(t) and not y(t) ? > >> >>>> 1:  The solution as you state the problem is trivial: y(t) = x(t). > >> >>>>  I assume you mean the frequency should be close to but different. > > >> >>>> 2:  Have you asked your prof? > > >> >>>> 3:  Read this:  http://www.wescottdesign.com/articles/Sampling/ > >> >>>> sampling.html.  Skip down to the part about aliasing. > >> >>> First, fix the link: > >> >>>http://www.wescottdesign.com/articles/Sampling/sampling.html > > >> >>> I'm puzzled by Tim's reference to aliasing. The frequency is 250 Hz > >> >>> and the sample rate is 1000 Hz. > > >> >>> Jerry > > >> >> He's asking for a continuous-time signal (presumably not identical > >> >> to the given one) that gives the same discrete-time signal after > >> >> sampling. > >> >>  That sounds like aliasing to me. > > >> >> It sounds like a _homework problem_ about aliasing.  One I might > >> >> write were I teaching a signal processing course, I might add. > > >> > Got it. I would have worded it differently, though. > > >> I would have worded it differently, too.  A set of homework problems > >> that doesn't contain at least one veiled reference to Bart Simpson, > >> Batman, Diogenes, Gilligan, or some other major philosopher is a minor > >> failure, IMHO. > > > Cool - can you reword the OPs problem in that vein? > > I've got it: > > Gilligan has sprung a net trap set by natives from the next island over; > the trap is hung from a tall tree branch, and Gilligan is swinging over a > small clearing with a period of oscillation of 5 seconds (0.2Hz). > > The Skipper has also been captured and blindfolded and is being held > nearby.  Every two seconds exactly he manages to pull off the blindfold > and see Gilligan, but his captors immediately replace the blindfold each > time. > > What is the _closest_ (but not identical) period of oscillation at which > Gilligan could be swinging such that the Skipper would see exactly the > same period of oscillation given how often he pulls off his blindfold? > > For extra credit: > > The Skipper escapes, and runs to the Professor to explain Gilligan's > dilemma (including the timing, of course).  The Professor (having taken > signal processing courses along with everything else), knows that there > are two likely periods to the swing, and thus knows the two possible > lengths of the rope from which Gilligan is suspended (he ignores the mass > of the rope and assumes the tree is infinitely rigid). > > Assuming that the professor gets his math right, how long are his two > calculated rope lengths? :-) Unfortunately, I don't know Gilligan. Wikipedia has an article about a sitcom called Gilligan's Island, which I guess you are refering to. Sounds like an older version of "Lost" ... |