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【简答题】
调频波的解调常用（）解调方式，鉴频器最重要的功能是（），使所得的解调信号具有（）的信息。
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参考答案：
举一反三
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
【单选题】Bitmap的可能长度为0，这个Bitmap第一位指示SN为（）ol4096的SDU是否被正确接收解调
A.
FMS+0
B.
FMS+1
C.
FMS+2
D.
FMS+3
相关题目：
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
【单选题】Bitmap的可能长度为0，这个Bitmap第一位指示SN为（）ol4096的SDU是否被正确接收解调
A.
FMS+0
B.
FMS+1
C.
FMS+2
D.
FMS+3
参考解析：
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