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== Part E: Linearity and Time Invariance == | == Part E: Linearity and Time Invariance == | ||
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| + | === Part A === | ||
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| + | For the output signal to be time invariant, the response to the shifted input signal <math> x[n-N] </math> should be the shifted output (<math> y[n-N] </math>). Basically, this means that if the input signal is shifted along the x-axis by any amount of time, the output signal should produce the same value at <math> n + N </math> that it used to produce at <math> n </math> before the shift. | ||
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| + | For the system given, let's use the signal/system corresponding to <math> X </math><sub>2</sub> for the first one and <math> X </math><sub>3</sub> for the second. Then select a time <math> n </math> of 4, and a shifted time <math> N </math> of 1. If this system was time invariant, <math> Y </math><sub>2</sub><math>[n] </math> at time <math> n = 5 </math> should equal <math> Y </math><sub>1</sub><math>[n] </math> at <math> n = 4 </math>. | ||
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| + | <math>\,\ X </math><sub>2</sub><math> \,\ [4] = \delta (2) </math><br> | ||
| + | <math>\,\ Y </math><sub>2</sub><math> \,\ [4] = 9\delta (1) </math> | ||
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| + | <math>\,\ X </math><sub>3</sub><math> \,\ [5] = \delta (2) </math><br> | ||
| + | <math>\,\ Y </math><sub>3</sub><math> \,\ [5] = 16\delta (1) </math> | ||
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| + | <math> 9\delta (1) \ne 16\delta (1) </math><br> | ||
| + | <math>\,\ Y </math><sub>2</sub><math> \,\ [4] \ne Y </math><sub>3</sub><math> \,\ [5] </math><br> | ||
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| + | Therefore, the system is '''not''' time invariant. | ||
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| + | === Part B === | ||
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| + | We are given the premise that <math> X[n] </math> is linear. Because this is given to us, we can use the Time Invariance property which states that the output response of a shifted input signal is the shifted output response. That means all that would need to be done to produce an output of <math> Y[n] = u[n-1] </math> would be to use an input signal of <math> X[n] = u[n] </math>. | ||
Latest revision as of 15:51, 11 September 2008
Part E: Linearity and Time Invariance
Part A
For the output signal to be time invariant, the response to the shifted input signal $ x[n-N] $ should be the shifted output ($ y[n-N] $). Basically, this means that if the input signal is shifted along the x-axis by any amount of time, the output signal should produce the same value at $ n + N $ that it used to produce at $ n $ before the shift.
For the system given, let's use the signal/system corresponding to $ X $2 for the first one and $ X $3 for the second. Then select a time $ n $ of 4, and a shifted time $ N $ of 1. If this system was time invariant, $ Y $2$ [n] $ at time $ n = 5 $ should equal $ Y $1$ [n] $ at $ n = 4 $.
$ \,\ X $2$ \,\ [4] = \delta (2) $
$ \,\ Y $2$ \,\ [4] = 9\delta (1) $
$ \,\ X $3$ \,\ [5] = \delta (2) $
$ \,\ Y $3$ \,\ [5] = 16\delta (1) $
$ 9\delta (1) \ne 16\delta (1) $
$ \,\ Y $2$ \,\ [4] \ne Y $3$ \,\ [5] $
Therefore, the system is not time invariant.
Part B
We are given the premise that $ X[n] $ is linear. Because this is given to us, we can use the Time Invariance property which states that the output response of a shifted input signal is the shifted output response. That means all that would need to be done to produce an output of $ Y[n] = u[n-1] $ would be to use an input signal of $ X[n] = u[n] $.
