Skip to main content

Behaviour of x[n] = cαⁿ

Short answer

For x[n]=cα^n with real c,α (c>0): α>1 grows unbounded; 0<α<1 decays; α=1 constant; -1<α<0 alternates sign and decays; α<-1 alternates and grows; α=-1 alternates ±c. A negative c flips the sequence.

Explain the behaviour of discrete time complex exponential x[n] = cαⁿ where c and α are real.

With cc and α\alpha both real, x[n]=cαnx[n] = c\alpha^n is a real exponential sequence. Its shape depends on the value of α\alpha (take c>0c > 0):

Value of α\alphaBehaviour of x[n]x[n]
α>1\alpha > 1Grows exponentially with nn (unbounded)
α=1\alpha = 1Constant, x[n]=cx[n] = c
0<α<10 < \alpha < 1Decays exponentially towards 0
−1<α<0-1 < \alpha < 0Alternates in sign and decays
α=−1\alpha = -1Alternates between +c+c and −c-c
α<−1\alpha < -1Alternates in sign and grows

For negative α\alpha, αn\alpha^n is positive for even nn and negative for odd nn, giving the alternating pattern. A negative cc simply flips the whole sequence.

 0 < alpha < 1: decays
  o
  |
  |  o
  |  |  o  o
--+--+--+--+--o-> n
  0  1  2  3  4

 alpha > 1: grows
              o
              |
           o  |
        o  |  |
  o  o  |  |  |
--+--+--+--+--+-> n
  0  1  2  3  4

 -1 < alpha < 0: alternates, decays
  o
  |
  |
  |     o
--+--+--+--+--o-> n
     |     o
     o
  0  1  2  3  4

 alpha < -1: alternates, grows
        o
        |
  o     |
--+--+--+--+-> n
     |     |
     o     |
           |
           o
  0  1  2  3

Examples: (0.5)nu[n](0.5)^n u[n] decays (used for stable system impulse responses); (−0.8)nu[n](-0.8)^n u[n] oscillates and decays; 2nu[n]2^n u[n] grows (unstable).

✓ Answer checked by Claude Opus 5.5 (high effort)

All Signals and Systems chapter 1 questions →