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For x(n)={2 3 4 5 1 3}, plot the following Discrete Time signals.

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For x(n)={2 3 4 5 1 3}, plot the following Discrete Time signals.

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written 6.0 years ago by |

1) x(n-1)

Solution: $n’ = n-1 \\ \therefore n=n’-1$

n | x(n) | n’ |
---|---|---|

-2 | 2 | -3 |

-1 | 3 | -2 |

0 | 4 | -1 |

1 | 5 | 0 |

2 | 1 | 1 |

3 | 3 | 2 |

2) x(n)u(-n)

Solution: $n’ = -n \\ u[n] = 0, n\lt0 \\ = 1, n≥0$

n | x(n) | u(n’) | x(n)u(-n) |
---|---|---|---|

-2 | 2 | 1 | 2 |

-1 | 3 | 1 | 3 |

0 | 4 | 1 | 4 |

1 | 5 | 0 | 0 |

2 | 1 | 0 | 0 |

3 | 3 | 0 | 0 |

3) x(n-1)u(-n-1)

Solution: $n’ = n-1 \\ ∴ n = n’+1$

4) x(-n)u(n)

Solution: n’ = -n

n | n’ | x(n’) | u(n’) | x(n’)u(n) |
---|---|---|---|---|

-2 | 2 | 2 | 1 | 2 |

-1 | 1 | 3 | 1 | 3 |

0 | 0 | 4 | 1 | 4 |

1 | -1 | 5 | 0 | 0 |

2 | -2 | 1 | 0 | 0 |

3 | -3 | 3 | 0 | 0 |

**Result: x(-n)u(n) = {4}**

5) x(2n)

Solution: $n’ = 2n \\ \therefore n = n’/2$

n | n’ | x(n’) |
---|---|---|

-2 | -1 | 2 |

-1 | -1/2 | 3 |

0 | 0 | 4 |

1 | 1/2 | 5 |

2 | 1 | 1 |

3 | 3/2 | 3 |

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