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Michael Renardy
Michael Renardy
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Yes, this is true. It follows from the fact that the average $$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum A_i\cos(\lambda_i t+\phi_i)\,dt$$$$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum_j A_j\cos(\lambda_j t+\phi_j)\,dt$$ is $A_i/2$. Clearly the average is bounded by the supremum of the function.

Yes, this is true. It follows from the fact that the average $$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum A_i\cos(\lambda_i t+\phi_i)\,dt$$ is $A_i/2$. Clearly the average is bounded by the supremum of the function.

Yes, this is true. It follows from the fact that the average $$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum_j A_j\cos(\lambda_j t+\phi_j)\,dt$$ is $A_i/2$. Clearly the average is bounded by the supremum of the function.

Source Link
Michael Renardy
Michael Renardy
  • 13k
  • 1
  • 42
  • 50

Yes, this is true. It follows from the fact that the average $$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum A_i\cos(\lambda_i t+\phi_i)\,dt$$ is $A_i/2$. Clearly the average is bounded by the supremum of the function.